1.1 Manifolds and Maps

The title of this course is “Introduction to Differential Manifolds,” which suggests that these differential manifolds (or sometimes differentiable manifolds), whatever they are, will probably be important. So what is a differential manifold? The name should suggest the answer: they are spaces in which we know what differentiation is supposed to mean. I actually prefer the term smooth manifold, so that is what I will use going forward, though this will often just get shortened to manifold.

In practice, the idea is to leverage the fact that we already (hopefully!) know how to do calculus in \(\mathbb {R}^n\) (this is exactly what MATH 261 is all about), and to translate those techniques to more general spaces. The key insight here is that differentiation is a local operation: to compute a derivative at a point (whether it’s a gradient, curl, divergence, directional derivative, whatever), you really only need to know what’s going on in a tiny open neighborhood of that point.

So to get a space on which we can compute derivatives, it’s enough to have a space which is “locally Euclidean” or “locally like \(\mathbb {R}^n\),” and this is what manifolds are. Roughly speaking, this means that around any point in a manifold you can find a small open set which looks just like an open set in some \(\mathbb {R}^n\), and then we can do calculus on the manifold by translating in a neighborhood of a point to the corresponding set in \(\mathbb {R}^n\), where we know what to do.

This is all to say that the point of defining manifolds in the way we are about to (which is extremely non-obvious and unintuitive!) is that these are precisely the spaces in which a suitable generalization of multivariable calculus makes sense.

So what does “locally like \(\mathbb {R}^n\)” actually mean? Here’s a standard definition:

See Figure 1.1 for a visualization of 2, showing the map \(\phi _\beta ^{-1} \circ \phi _\alpha \) from \(\phi _\alpha ^{-1}(W) \subseteq \mathbb {R}^n\) to \(\phi _\beta ^{-1}(W) \subseteq \mathbb {R}^n\).

Of course, most interesting manifolds are not \(\mathbb {R}^n\), but the idea of 1 from Definition 1.1.1 is that you can cover any manifold by a bunch of little open sets (namely, the \(\phi _\alpha (U_\alpha )\)) that are essentially identical to open sets in \(\mathbb {R}^n\) (namely, the \(U_\alpha \)), ² so you can essentially do any local calculation in \(\mathbb {R}^n\). It’s also very important that the \(n\) is always the same here: if \(m \neq n\), we’re not allowed to have some points with neighborhoods that look like \(\mathbb {R}^m\) and some other points whose neighborhoods look like \(\mathbb {R}^n\).

If the idea is to use the coordinate charts to transport calculations from the manifold to \(\mathbb {R}^n\), then a thing you should be very worried about is that, if a point lies in two different charts, then there are two different ways to do this and they might not be compatible. This is the point of 2: whether you do your calculations in \(U_\alpha \) or \(U_\beta \), the two are related by a smooth map, so you can easily translate between the two calculations using the change-of-variables formula. Indeed, the usual English-language gloss of 2 is that “transition functions are smooth.”

Finally, 3 is a technical condition that in practice is not important. The point of it is simply to ensure uniqueness: if you had a collection of coordinate charts satisfying 1 and 2, and I took your collection and added some new charts while still satisfying 2, it would be kind of silly to say that you and I were talking about different manifolds. Taking maximal families gives uniqueness since your collection of charts and my collection of charts live in the same maximal family.

However, it is certainly possible to have distinct maximal collections on the same space (if there is one that is in some sense standard, then any others are sometimes called “exotic smooth structures”). At least two Fields Medals have been awarded primarily for finding examples of exotic smooth structures: to John Milnor in 1962 (for finding exotic 7-spheres  [ 24 ] ; it turns out there are exactly 28 distinct smooth structures on \(S^7\)  [ 20 ] ) and to Simon Donaldson in 1986 (for finding exotic \(\mathbb {R}^4\)s  [ 10 , 12 , 16 ] ; it turns out there are uncountably many distinct smooth structures on \(\mathbb {R}^4\)  [ 29 ] ). It remains an open problem called the smooth 4-dimensional Poincaré conjecture whether there are non-standard differentiable structures on \(S^4\).

Example 1.1.5

\(S^n\) the unit sphere in \(\mathbb {R}^{n+1}\) is a manifold. Specifically, I claim that the maximal family containing \(\{ (R^n, \phi _N), (R^n, \phi _S)\} \) makes \(S^n\) into an \(n\)-dimensional manifold, where \(\phi _N\) and \(\phi _S\) are inverse stereographic projection from the north and south poles, respectively.

Specifically, with \(\vec{x} = (x_1, \dots , x_n) \in \mathbb {R}^n\), define

\[ \phi _N(\vec{x}) := \frac{1}{1+\| \vec{x}\| ^2}\left(2x_1, \dots , 2x_n, -1+\| \vec{x}\| ^2\right) \]

and

\[ \phi _S(\vec{x}) := \frac{1}{1+\| \vec{x}\| ^2}\left(2x_1, \dots , 2x_n, 1-\| \vec{x}\| ^2\right). \]

Then \(\phi _N\) is the map that sends \(\vec{x} \in \mathbb {R}^n\) to the point on the sphere which lies on the line segment connecting \((x_1,\dots , x_n, 0) \in \mathbb {R}^{n+1}\) to the north pole \((0, \dots , 0, 1) \in S^n \subset \mathbb {R}^{n+1}\); see Figure 1.2.

To prove the claim, we need to show that 1 and 2 from Definition 1.1.1 are satisfied (3 is automatically satisfied, since we’re taking the maximal family containing \(\{ (\mathbb {R}^n, \phi _N), (\mathbb {R}^n, \phi _S)\} \)).

If we let \(N = (0,\dots , 0, 1)\) be the north pole and \(S = (0,\dots , 0, -1)\) the south pole, then \(\phi _N(\mathbb {R}^n) = S^n \backslash \{ N\} \) and \(\phi _S(\mathbb {R}^n) = S^n \backslash \{ S\} \) and the union is all of \(S^n\), so 1 is satisfied.

For 2, observe that

\[ W = \phi _N(\mathbb {R}^n) \cap \phi _S(\mathbb {R}^n) = S^n \backslash \{ N,S\} , \]

so

\[ \phi _N^{-1}(W) = \mathbb {R}^n \backslash \{ 0\} , \]

which is certainly open, and likewise for \(\phi _S^{-1}(W) = \mathbb {R}^n \backslash \{ 0\} \). So we need to verify that \(\phi _N^{-1} \circ \phi _S\) and \(\phi _S^{-1} \circ \phi _N\) are smooth as functions on \(\mathbb {R}^n \backslash \{ 0\} \).

The inverse of \(\phi _N\) is stereographic projection

\[ \phi _N^{-1}(\vec{y}) := \frac{1}{1-y_{n+1}} (y_1, \dots , y_n) \]

(check this!) so we see that

\[ (\phi _N^{-1} \circ \phi _S)(\vec{x}) = \frac{1}{\| \vec{x}\| ^2}\vec{x} \]

is reflection through the unit sphere in \(\mathbb {R}^n\), which is smooth away from the origin. And similarly for \(\phi _S^{-1} \circ \phi _N\).

As already mentioned, the point of manifolds is that they are spaces in which we can do calculus, so we should be able to say what it means for a map between manifolds to be differentiable. Hopefully it’s already starting to become clear what the strategy is: we can talk about differentiability at a point, and then both the point in the domain and the point it maps to in the range lie in coordinate charts that are like open sets in Euclidean spaces. So then locally our map just looks like a map between Euclidean spaces, where we already know what it means for a map to be differentiable.

While we’ve given the definition of a manifold and of a differentiable map in this section, we generally try to use them directly as little as possible. They are hard to handle and fairly unintuitive, so we will quickly be looking for alternative ways of characterizing manifolds and differentiable maps.

1.2 Tangent Vectors

Notice that there’s nothing in Definition 1.1.1 that says that a manifold has to live inside some bigger Euclidean space. This is in contrast to a typical undergraduate differential geometry course (like MATH 474 here at CSU), which is typically focused on surfaces in \(\mathbb {R}^3\).

Of course, many manifolds (like spheres) do naturally live in some Euclidean space, and it turns out that the Whitney embedding theorem  [ 30 , 31 ] guarantees that all manifolds can be embedded in some Euclidean space, but this embedding is not necessarily going to be pleasant to work with. If you’ve encountered them before, it is often much easier to work with projective spaces and Grassmannians without embedding them anywhere in particular.

Since our manifolds don’t necessarily live inside a Euclidean space, we have to be a bit careful about what a tangent vector at a point is supposed to be. In particular, a tangent vector at a point lives in a different universe than the point itself: the point is a point in the manifold, but the tangent vector does not live on the manifold. This is clear even on a sphere in Euclidean space: the tangent vector to a point on the sphere does not live on the sphere! However, in Euclidean space we can kind of cheat and think of both the point and the tangent vector as both living inside the ambient Euclidean space. In general, we can’t get away with this.

Now even in Euclidean space you have to be a little careful with this mixing of point and tangent vector: the tangent vector can’t be any arbitrary vector in the ambient Euclidean space: it has to lie in the tangent space at the point, which we usually visualize as some plane which is tangent to the sphere at the point. But if we’re not in Euclidean space, things are even worse: if there’s supposed to be some subspace which is “tangent at a point,” where does it even live? Surely not in the manifold itself, but we’re not thinking of the manifold as sitting inside some bigger space, so there’s no “outside” where it can be.

This is a surprisingly nontrivial issue, requiring us to construct some abstract vector space which doesn’t really live anywhere in particular. The construction is fairly non-obvious, and feels like a sneaky trick the first few times you encounter it. This is one of those situations where it seems like you’re turning a concept inside-out; at least for me, the first time I encountered the following way of thinking, it made me feel uncomfortable in a similar way to when I first encountered the natural embedding of a vector space into its double dual. I will say that at some point my brain switched from “this is weird and awkward” to “this is obviously the right way to do it” and I think most differential geometers have had similar experiences, so this is something you can eventually develop intuition for.

Given that preamble, what’s the idea? We’re going to work by analogy with a way of thinking about vectors in \(\mathbb {R}^n\) that’s slightly different from what you may be used to. In words, we’ll identify a vector \(\vec{v} \in \mathbb {R}^n\) with the operator on differentiable functions which gives the directional derivative (in the direction of \(\vec{v}\)) of a function.

That’s a little vague, so let’s try to characterize a tangent vector \(\vec{v}\) to a point \(p \in \mathbb {R}^n\) in this way (here I’m using the notation \(p\) rather than \(\vec{p}\), because I’m just thinking of \(p\) as a point in a manifold, not as an element of a vector space). Given \(\vec{v}\), I claim we can find some smooth curve \(\alpha \colon \thinspace \! (-\epsilon , \epsilon ) \to \mathbb {R}^n\) with \(\alpha (0) = p\) and \(\alpha '(0) = \vec{v}\); see Figure 1.4. In coordinates, if

where the coordinate functions \(x_i\colon \thinspace \! (-\epsilon ,\epsilon ) \to \mathbb {R}\) are themselves smooth, then

(In this specific case, we could take \(\alpha (t) = p + t\vec{v}\), but it turns out not to matter which curve satisfying \(\alpha (0) = p\) and \(\alpha '(0) = \vec{v}\) we take.)

Now, say \(f: U \to \mathbb {R}\) is differentiable, where \(U \subseteq \mathbb {R}^n\) is some neighborhood of \(p\). Then the directional derivative of \(f\) at \(p\) in the direction of \(\vec{v}\) is

by the Chain Rule.

On the right hand side above, we’ve written the directional derivative as an operator \(L = \sum _{i=1}^n x_i'(0) \frac{\partial }{\partial x_i}\) acting on \(f\). Moreover, this operator depends uniquely on \(\vec{v}\) and is a linear derivation, meaning that:

1. \(L(f + \lambda g) = L(f) + \lambda L(g)\) for all \(f\) and \(g\) differentiable in a neighborhood of \(p\) and all \(\lambda \in \mathbb {R}\);

2. \(L(f g) = f(p) L(g) + g(p) L(f)\) (i.e., the Product Rule).

If it’s not clear to you, it’s worth looking at the above and convincing yourself that the operator \(L\) didn’t really depend on the choice of \(\alpha \): it really only depends on \(p\) and \(\vec{v}\) (you may need to go back to the justification from MATH 517 that the directional derivative is well-defined).

The upshot is that, given a point \(p\) and a tangent vector \(\vec{v}\) at \(p\), we get a directional derivative operator \(L\). And, conversely, if we know how to compute a directional derivative, we (at least implicitly) know the direction, so this is really a bijective correspondence.

The benefit of thinking in this way is that we can talk about differential operators like \(L\) on any manifold; after all, manifolds locally look like \(\mathbb {R}^n\) by definition. So, with that long preamble in mind, here’s the definition of a tangent vector:

This is a nice coordinate-free way of defining things, but it’s not very useful for computations. We usually do want to work in coordinates for computations (and certainly for any computations that we want to do on the computer), so let’s see what all this means in local coordinates.

Say that \((U,\phi )\) is a coordinate chart containing \(p \in M\) so that \(\phi (\vec{0}) = p\), that \(\alpha \colon \thinspace \! (-\epsilon , \epsilon ) \to M\) is smooth with \(\alpha (0) = p\), and that \(f\) is a differentiable function in a neighborhood of \(p\). Then

for some smooth functions \(x_1 , \dots , x_n\colon \thinspace \! (-\epsilon , \epsilon ) \to \mathbb {R}\). Then

where I’m using a very common abuse of notation in the second expression to think of \(f\) as being a function on the coordinate chart \(U\), and hence a function of coordinates \(x_1, \dots , x_n\) (of course, it’s really \(f \circ \phi \) which is a function of \(x_1, \dots , x_n\), as we see in the middle expression).

This all means that we can write the tangent vector \(\alpha '(0) \in T_pM\) in local coordinates as

Since the \(x_i'(0)\) are just scalars, what we’re doing here is writing \(\alpha '(0)\) in terms of the local coordinate basis \(\left\{ \left(\frac{\partial }{\partial x_1}\right)_{\vec{0}}, \dots , \left(\frac{\partial }{\partial x_n}\right)_{\vec{0}}\right\} \) for \(T_pM\) associated to the chart \((U,\phi )\).

It’s extremely important to keep in mind that, if \(p\) and \(q\) are distinct points on \(M\), then the tangent spaces \(T_pM\) and \(T_qM\) are completely different vector spaces that in principle have nothing to do with each other (they’re both vector spaces of the same dimension, and hence abstractly isomorphic, but that’s essentially all we can know without more detailed information about the geometry of \(M\)).

Nonetheless, we often want to talk about all tangent spaces at once, so we smush them together:

Notice that there are natural projections \(\pi : TM \to M\) and \(\widetilde{\pi }: T^\ast M \to M\) which just record the base point; in other words, \(\pi \) sends a tangent vector at a point to the point (formally, \(TM\) and \(T^\ast M\) are vector bundles, and this projection is part of their definition).

\(T^\ast M\) is, in some sense, the most basic example of a symplectic manifold. In physics and dynamical systems language, if \(M\) is the configuration space ³ of a (classical) system, then \(T^\ast M\) is phase space or position-momentum space: this is the natural setting of Hamiltonian mechanics.

1.3 The Differential

In multivariable calculus, any differentiable map \(f \colon \thinspace \! \mathbb {R}^m \to \mathbb {R}^n\) has a corresponding differential \(df\). ⁴ While in a multivariable calculus setting we often think of \(df\) as a map \(\mathbb {R}^m \to \mathbb {R}^n\) as well (meaning it would have the same domain and codomain as \(f\)), or even more concretely as an \(n \times m\) matrix, this is misleading.

In fact, if you were reading the first sentence of the previous paragraph very carefully, you may have noticed that it was not quite correct. Really, I should have said that any differentiable map has a corresponding differential at a point \(p\). There is no single “differential” for a non-linear map: we get different differentials in the neighborhood of each point in the domain.

So more accurately, if \(p\) is in the domain of \(f\), then we have a differential \(df_p\) at \(p\). Now, what is the domain of \(df_p\)? Recall, \(df_p\) is supposed to be the best linear approximation of \(f\) at \(p\), so we’re supposed to think of the inputs to \(df_p\) as being slight perturbations of \(p\). In other words, the domain of \(df_p\) is really the tangent space \(T_p \mathbb {R}^m\) (which is in a very precise sense the space of infinitesimal perturbations of \(p\)).

Now, \(T_p \mathbb {R}^m\) is isomorphic to \(\mathbb {R}^m\), but not just in some abstract sense: the tangent bundle \(T\mathbb {R}^m\) is trivial, meaning that \(T \mathbb {R}^m \cong \mathbb {R}^m \times \mathbb {R}^m\), and a concrete trivialization is given by parallel translating a tangent vector from being based at \(p\) to being based at the origin. This is the sense in which we can think of the domain of each \(df_p\) as being the “same” \(\mathbb {R}^m\).

But if \(M\) is some more general \(m\)-dimensional manifold, then, while any tangent space \(T_p M\) is still abstractly isomorphic to \(\mathbb {R}^m\), the tangent bundle is likely not to be trivial, so we can’t make any general identifications of different tangent spaces with the same \(\mathbb {R}^m\). So if we want to generalize the notion of differential to manifolds, we need to be careful, even when thinking about functions on \(\mathbb {R}^m\), to maintain the distinction between \(\mathbb {R}^m\) (thought of an \(m\)-dimensional manifold) and \(T_p \mathbb {R}^m\).

And what about the codomain of \(df_p\)? Again, we should think more conceptually about what the differential really does. Like all derivatives, the differential tells us something about how the change in an input to a function changes the output of the function. So the inputs to the differential will be tangent vectors at some point in the domain of \(f\) (recording an infinitesimal change to the input, or equivalently an initial position [the point] and velocity [the tangent vector] of some path), and the outputs will be tangent vectors at the corresponding point in the range (recording the infinitesimal change in the output). Notice that \(T_q \mathbb {R}^n\) is again isomorphic with \(\mathbb {R}^n\) by parallel translation.

In other words, given a differentiable map \(f \colon \thinspace \! \mathbb {R}^m \to \mathbb {R}^n\) and a point \(p\) in the domain of \(f\), the differential of \(f\) at \(p\) is a map \(df_p \colon \thinspace \! T_p \mathbb {R}^m \to T_{f(p)}\mathbb {R}^n\), where domain and range are usually identified with \(\mathbb {R}^m\) and \(\mathbb {R}^n\) in a standard way. This, then, is the way of thinking about differentials which generalizes.

In the special case that \(f\colon \thinspace \! M \to \mathbb {R}\) is a smooth function, the differential of \(f\) applied to a tangent vector \(v\) is the same thing as computing the derivative of \(f\) in the direction of \(v\):

But this is exactly the definition of \(df_p v\).

Let’s see what this means in local coordinates, which is how we’ll actually compute in practice. Suppose \(p \in M\) and \((V, \psi )\) is a local coordinate chart on \(N\) containing \(f(p)\). By Definition 1.1.6, there exists some compatible chart \((U, \phi )\) on \(M\) containing \(p\) so that \(f(\phi (U)) \subseteq \psi (V)\). Then the curve \(\alpha \) on \(M\) (or, really, its restriction to \(\phi (U)\)) gives a corresponding curve

in \(U \subseteq \mathbb {R}^m\), and similarly the curve \(\beta \) on \(N\) has a corresponding curve

See Figure 1.5.

Of course,

so we can think of the \(y_i\) as functions of the \(x_j\), and the curve in \(V\) can be written as

Working in local coordinates means doing the computation at the level of the map \(\psi ^{-1} \circ f \circ \phi \colon \thinspace \! U \subseteq \mathbb {R}^m \to V \subseteq \mathbb {R}^n\), which has a corresponding differential

By definition, evaluating this differential on the tangent vector \((x_1'(0), \dots , x_m'(0)) \in T_{(x_1(0),\dots , x_m(0))}\mathbb {R}^m\) gives

where \(\begin{bmatrix} \frac{\partial y_i}{\partial x_j} \end{bmatrix}_{i,j} \) is the \(n \times m\) matrix of partials (i.e., the Jacobian).

So when we work in local coordinates, this is just standard multivariable calculus, as you would hope.

Just to reiterate, the differential (at a point) of a map between manifolds is going to input a tangent vector at a point in the domain and output a tangent vector at a point in the range. As a special case, suppose we have a smooth function on a manifold \(M\); that is, a smooth map \(f \colon \thinspace \! M \to \mathbb {R}\). Then, at a point \(p \in M\), the differential \(df_p \colon \thinspace \! T_p M \to T_{f(p)}\mathbb {R}\). Now, the tangent bundle to \(\mathbb {R}\) is trivial, so we can identify \(T_{f(p)}\mathbb {R}\) with \(\mathbb {R}\) in a canonical way, and therefore think about \(df_p\) as a linear map \(T_pM \to \mathbb {R}\).

In other words, by way of this identification of \(T_{f(p)} \mathbb {R}\) with \(\mathbb {R}\), we can think of \(df_p\) as a linear functional on the vector space \(T_pM\). Or, in fancier language, \(df_p\) is an element of the dual space \(\left(T_pM\right)^\ast \). We’ll come back to this later when we start talking about differential forms, but this gives a hint of why we might care about cotangent spaces and the cotangent bundle.

Just as in multivariable calculus, when we have a smooth map \(f \colon \thinspace \! M \to N\), we can talk about critical points and critical values, and doing so is often quite important.

Example 1.3.6 (Cliché Example)

Consider the function \(f\) on the torus depicted in Figure 1.6. In words, \(f\) is the height function on this upright torus that just records the \(z\)-coordinate. I’ve shown the critical values in red, as well as a couple of regular values in green. Also, back on the torus I’ve shown the inverse images of the critical and regular values, as well as some tangent vectors to regular points and their images under the differential (in blue and purple), to hopefully convince you the differential really is surjective at these points.

Notice that there are basically three different types of critical points: a minimum, a maximum, and two saddle points. The differential can’t possibly be surjective at a (local) minimum: there’s no direction you could go which will cause the function to decrease, so the image of the differential has to miss an entire half of its codomain (and therefore, by linearity, everything except the origin). It’s somewhat less geometrically clear that the saddle points are critical points, though in this case it’s fairly easy to see once you realize the tangent plane to one of the saddle points is horizontal.

A couple of other things to notice about this picture:

1. The inverse images of the two critical points coming from the saddle points contain plenty of regular points: indeed, the differential is surjective at any of the points in the inverse image except the saddle point.

2. The level sets of regular values are smooth (collections of) curves, either a single circle or two disjoint circles; in other words, they are smooth (possibly disconnected) 1-dimensional manifolds. The level sets of the critical points are either a single point (the minimum and maximum) or a wedge of two circles (like an \(\infty \) symbol). In both cases, the result is not a smooth 1-manifold: a point is a 0-manifold and the wedge of two circles is not locally like a copy of \(\mathbb {R}\) near the point where the two circles meet (which of course is exactly the saddle point).

Example 1.3.8

Consider the function \(f(x,y,z) = z^2\) on the sphere (see Figure 1.7). Again, I’ve shown critical values in red and orange and a regular value in green, and the corresponding level sets on the domain, as well as some tangent vectors to regular points and their images under the differential.

Some new features:

1. The critical points aren’t all isolated: the entire equator consists of critical points (which are all global minima).

2. The inverse images of critical values aren’t necessarily connected: the north and south poles both map to 1.

Notice again that the level sets of regular values are (collections of) smooth curves. In this case one critical level set is also a smooth curve, and the other is not a curve at all.

(This is not a Morse function because not all critical points are isolated, but it is a Morse–Bott function, which is almost as good.)

After looking at these examples, hopefully the following theorem suggests itself:

This theorem is basically an application of the Inverse Function Theorem, but before we work on proving it, let’s look at some more examples.

1.4 Immersions and Embeddings

Let’s work up to proving Theorem 1.3.9, including defining some more terminology.

Example 1.4.2

Define \(\alpha \colon \thinspace \! \mathbb {R}\to \mathbb {R}^2\) by \(\alpha (t) = (t^3, t^2)\) (see Figure 1.8).

This is not an immersion because \(d\alpha _0 = \begin{bmatrix} \alpha _1'(0) \\ \alpha _2'(0) \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\) has rank 0. Intuitively, the problem is that the velocity is zero when \(t=0\), even though the map overall is continuous and injective.

Example 1.4.3

Define \(\beta \colon \thinspace \! \mathbb {R}\to \mathbb {R}^2\) by \(\beta (t) = (t^3 - 4t, t^2 - 4)\), which is a deformation of the previous example (see Figure 1.9).

Now the differential is given by

\[ d\beta _t = \begin{bmatrix} 3t^2 - 4 \\ 2t \end{bmatrix}, \]

which is never the zero matrix since the second entry being 0 implies \(t=0\), and hence that the first entry is \(-4\). Hence, \(d\beta _t\) is always rank 1, and hence always injective, so \(\beta \) is an immersion.

However, \(\beta \) is not an embedding because it is not injective: \(\beta (-2) = (0,0) = \beta (2)\), so there is a double point at the origin.

If \(f \colon \thinspace \! M \to N\) is a local diffeomorphism at \(p \in M\), then \(df_p \colon \thinspace \! T_pM \to T_{f(p)}N\) is a linear isomorphism (i.e., invertible linear map). Perhaps somewhat surprisingly, the converse is also true. This is the appropriate generalization of the Inverse Function Theorem to the manifold setting, and the proof essentially involves applying the Inverse Function Theorem in local coordinates:

Notice that, by the Chain Rule,

But then we already know that \(d\phi \) and \(d\psi ^{-1}\) are isomorphisms (since coordinate charts are local diffeomorphisms [Example 1.4.4]), so \(df_p\) being an isomorphism implies that \(d(\psi ^{-1} \circ f \circ \phi )_{\phi ^{-1}(p)}\) is also an isomorphism.

But then the Inverse Function Theorem implies \(\psi ^{-1} \circ f \circ \phi \) is a local diffeomorphism at \(\phi ^{-1}(p)\). Since \(\psi \) and \(\phi ^{-1}\) are local diffeomorphisms (again, Example 1.4.4) and compositions of local diffeomorphisms are local diffeomorphisms, it follows that

is also a local diffeomorphism.

Here’s a statement of the Inverse Function Theorem in the language of differentials and local diffeomorphisms. It is equivalent to the usual statement you would see in a multivariable calculus or analysis course.

We’re now ready to prove the Level Set Theorem (Theorem 1.3.9):

Suppose \((U,\phi )\) is a coordinate chart centered at \(p \in f^{-1}(q)\) ⁵ and \((V, \psi )\) is a chart at \(q\). See Figure 1.10. Then define the map

By assumption,

is surjective since \(df_p\) is and the other two terms are isomorphisms.

Hence, after a linear change of coordinates \(dg_{\vec{0}}\) can be written as the \(n \times m\) block matrix \(\begin{bmatrix} I_{n \times n} & 0_{n \times (m-n)} \end{bmatrix}\). Define

which has differential \(dG_{\vec{0}} = I_{m \times m}\) in these coordinates. By the Inverse Function Theorem, \(G\) is a local diffeomorphism at \(\vec{0}\) and by construction \(g \circ G^{-1}\) is the standard projection onto the first \(n\) coordinates.

So in these coordinates \(f^{-1}(q) \cap \phi (U) = \phi (0, \dots , 0, x_{n+1}, \dots , x_m)\), so \(x_{n+1} , \dots , x_m\) give local coordinates near \(p\) for \(f^{-1}(q)\). We can do the same at any other point of \(f^{-1}(q)\), so this gives a system of coordinate charts on \(f^{-1}(q)\), and the transition maps on overlapping charts are smooth because they are just restrictions (or projections, if you prefer) of the corresponding transition maps for the charts on \(M\).

1.5 Vector Fields

Now that we have defined tangent vectors and seen how to push them around with differentials, the next natural object to try to define is a vector field. In physical problems, a vector field might be the velocity field of a fluid flow, or an electrical or magnetic field. In both applied and pure problems, we are very often interested in vector fields that arise as the gradients of functions, whether they be Morse functions on an abstract manifold or the energy of conformations in some conformation space. In symplectic geometry we are very interested in Hamiltonian vector fields associated to functions, which are a sort of symplectic gradient. Whereas the gradient is perpendicular to level sets of the function, Hamiltonian vector fields are parallel to level sets, so the function is constant on orbits, which is desirable when the function is an energy function.

So what is a vector field? Intuitively, it’s just what you would expect: a choice of tangent vector at each point in the manifold. As with everything in this class, we’re mostly interested in smooth vector fields, meaning that the tangent vector should in some sense vary smoothly as you move around the manifold.

The preceding sentence hopefully makes sense on an intuitive level, but you should stop and think about how you might try to make it into a rigorous definition. Unless you’ve already seen the forthcoming definition before, it’s probably not so easy.

The problem is that, as alluded to previously, different tangent spaces are not directly comparable. How do you compare \(v_1 \in T_{p_1}M\) with \(v_2 \in T_{p_2}M\) and, ideally, put them into some difference quotient?

Since we know how to compare tangent spaces in Euclidean space (by parallel translating), one strategy is to work in local coordinates and require the tangent vectors to form a smooth vector field in each local coordinate chart. But this rather unwieldy, and in any case we want to state definitions without reference to local coordinates if at all possible. Hence the following definition, which encapsulates exactly the idea above in a very concise (though admittedly kind of abstract and hard to visualize) way:

This requires some unpacking, especially since there’s at least one term in this definition which has not been defined yet.

First, recall (Theorem 1.2.5) that \(TM\) is a smooth manifold, so it makes sense to talk about smoothness of a map \(X\colon \thinspace \! M \to TM\). So the first part of Definition 1.5.1 is that a vector field is such a smooth map.

This makes sense: such a map takes as input a point in the manifold and outputs a tangent vector. But it would be nonsensical to output a tangent vector at \(q\) if the input is \(p\): the word “section” is how we rule this out.

More precisely, recall that we have a natural projection map \(\pi \colon \thinspace \! TM \to M\) which maps \(v \in T_p M\) to the base point \(p\). In general, if \(E \stackrel{\pi }{\to }B\) is a vector bundle, then a section of the bundle is a map \(\sigma \colon \thinspace \! B \to E\) so that \(\pi \circ \sigma = \operatorname {id}_B\), the identity map on \(B\) (in algebraic terms, \(\sigma \) is a right inverse of the projection \(\pi \)).

So when we say that \(X\) is a smooth section of the tangent bundle, we mean that \(X \colon \thinspace \! M \to TM\) is a smooth map satisfying \(\pi \circ X = \operatorname {id}_M\). In other words, that \(X\) assigns each point in \(M\) a tangent vector at that point in a smoothly-varying way. Which is exactly what a (smooth) vector field should be!

1.6 The Lie Bracket on Vector Fields

As we’ve defined it, \(\mathfrak {X}(M)\) is a \(C^\infty (M)\)-module: ⁶ we can certainly add two vector fields or multiply a vector field by a smooth function and get a new vector field.

In fact, there is even more algebraic structure on \(\mathfrak {X}(M)\): it is something called a Lie algebra, which means it is a (real) vector space which admits a binary operation (called a Lie bracket) satisfying certain axioms. ⁷

In order to define the Lie bracket of vector fields, it is instructive to think about how we could possibly define a binary operation on vector fields. For example, we could try to generalize binary operations on vector fields we already know in certain special cases. For example:

• Since \(T\mathbb {R}\cong \mathbb {R}\times \mathbb {R}\), we can interpret a vector field on \(\mathbb {R}\) as a real-valued function by just recording the second entry: \(X(p) = (p,v)\), and the corresponding function is \(p \mapsto v\). Since functions form an algebra by pointwise multiplication, this defines a binary operation on vector fields on \(\mathbb {R}\).

• More generally, every smooth 1-manifold has a trivial tangent bundle, so we can treat any vector field on any 1-manifold as a smooth, real-valued function and get an algebra structure on vector fields. ⁸

• Since \(T \mathbb {R}^2 \cong \mathbb {R}^2 \times \mathbb {R}^2\) and since we can define an equivalence \(\mathbb {R}^2 \leftrightarrow \mathbb {C}\) by \((x,y) \leftrightarrow x + iy\), we can interpret a vector field on \(\mathbb {R}^2\) (or, more generally, any surface with trivial tangent bundle) as a complex-valued function. Again, functions form an algebra by pointwise multiplication, so this defines a binary operation on vector fields on \(\mathbb {R}^2\). ⁹

• Since \(T\mathbb {R}^3 \cong \mathbb {R}^3 \times \mathbb {R}^3\), we can interpret a vector field on \(\mathbb {R}^3\) as a vector-valued function on \(\mathbb {R}^3\). Then the (pointwise) cross product \(\times \) gives a binary operation on vector fields on \(\mathbb {R}^3\).

• Since \(T\mathbb {R}^4 \cong \mathbb {R}^4 \times \mathbb {R}^4\) and since we can define an equivalence \(\mathbb {R}^4 \leftrightarrow \mathbb {H}\) between \(\mathbb {R}^4\) and the quaternions by \((t,x,y,z) \leftrightarrow t +i x + j y + kz\), we can interpret vector fields on \(\mathbb {R}^4\) as quaternion-valued functions. Again, functions form an algebra by pointwise multiplication, so this defines a binary operation on vector fields on \(\mathbb {R}^4\).

  This is in some sense a generalization of the cross product on \(\mathbb {R}^3\): if we represent \((x,y,z),(u,v,w) \in \mathbb {R}^3\) by purely imaginary quaternions \(p = ix + jy + kz\) and \(q = iu + jv + kw\), then the quaternion product

  \[ pq = -p \cdot q + p \times q, \]

  where \(p \cdot q\) is the dot product of the two vectors in \(\mathbb {R}^3\), and \(p \times q\) is the cross product (interpreted as a purerly imaginary quaternion). ¹⁰

So the natural question is: can we generalize these examples to higher-dimensional Euclidean spaces, and more generally to arbitrary manifolds?

Unfortunately, the answer is basically “no.” As you might have heard, there are no normed division rings over \(\mathbb {R}\) besides \(\mathbb {R}\), \(\mathbb {C}\), and \(\mathbb {H}\), and there aren’t even normed division algebras (where we allow multiplication to be non-associative) aside from these examples and the octonions \(\mathbb {O}\), which are 8-dimensional.

Moreover, even if there were more normed division algebras, the operations above depended very strongly on the triviality of the tangent bundle, which is not going to work in more general manifolds.

So here’s a different approach: recall that we can interpret vector fields on \(M\) as operators on \(C^\infty (M)\). That is, for \(f \in C^\infty (M)\), \(Xf \in C^\infty (M)\) as well; this is basically the function recording the directional derivative of \(f\) in the direction of \(X\) at each point on the manifold.

An obvious thing to guess is that, if \(X,Y \in \mathfrak {X}(M)\), then the composition \(X \circ Y = XY\) is also a vector field. After all, \((XY)f = X(Yf)\) will be a smooth function, so \(XY\) is also an operator on \(C^\infty (M)\).

Let’s work in local coordinates at a point \(p \in M\) to see what this operator looks like when applied to a function \(f\) which is differentiable at \(p\).

We know that, in terms of the local coordinate basis \(\left\{ \frac{\partial }{\partial x_1},\dots , \frac{\partial }{\partial x_n}\right\} \), we can write

where the \(a_i\) and \(b_i\) are smooth functions defined in some neighborhood of \(p\). So then

This should give you pause because it is no longer a first-order differential operator. Let’s make this more obvious by rewriting as

so the stuff inside the brackets (which is \(XY\)) is a differential operator on \(C^\infty (M)\), but it’s not a vector field; for example, it cannot be written in terms of the local coordinate basis \(\left\{ \frac{\partial }{\partial x_1}, \dots , \frac{\partial }{\partial x_n}\right\} \).

The problem with (1.1) is the second term, which is a second-order operator. Now notice that if we had done this in the reverse order we would have gotten

This doesn’t just have the same type of problem as before, it has literally the same problem: because mixed partials commute, the second terms in (1.1) and (1.2) agree. So, by subtracting, we can cancel them and get

which is written in terms of the \(\left\{ \frac{\partial }{\partial x_1}, \dots , \frac{\partial }{\partial x_n}\right\} \) basis. So this really is a vector field.

The key feature of this operation is that it satisfies the axioms of a Lie bracket (1–3 in Proposition 1.6.4), and hence makes \(\mathfrak {X}(M)\) into a Lie algebra.

Example 1.6.6

Consider \(M = S^2\) and the vector fields \(X = \frac{\partial }{\partial \theta }\) and \(Y = \frac{\partial }{\partial z}\) where \((\theta ,z)\) are cylindrical coordinates on \(S^2\); see Figure 1.11.

  

In other words, we have the local coordinate chart \(\phi \colon \thinspace \! (0,2\pi ) \times (-1,1) \to S^2\) given by

\[ \phi (\theta ,z) = (\sqrt{1-z^2} \cos \theta , \sqrt{1-z^2} \sin \theta , z), \]

and \(X\) and \(Y\) are the corresponding coordinate fields.

Notice that

\[ [X,Y]f = \frac{\partial ^2 f}{\partial \theta \partial z} - \frac{\partial ^2 f}{\partial z \partial \theta } = 0 \]

since mixed partials commute.

More generally, whenever \(X = \frac{\partial }{\partial x_i}\) and \(Y = \frac{\partial }{\partial x_j}\) are coordinate fields in a neighborhood of a point in a manifold, \([X,Y] = 0\).

In Cartesian coordinates

\[ X = \sqrt{1-z^2}\left( - \sin \theta \frac{\partial }{\partial x} + \cos \theta \frac{\partial }{\partial y}\right) = \sqrt{1-z^2}\left( - y \frac{\partial }{\partial x} + x \frac{\partial }{\partial y}\right), \]

and

\[ Y = -\frac{z \cos \theta }{\sqrt{1-z^2}}\frac{\partial }{\partial x} - \frac{z \sin \theta }{\sqrt{1-z^2}}\frac{\partial }{\partial y} + \frac{\partial }{\partial z} = -\frac{xz}{1-z^2}\frac{\partial }{\partial x} - \frac{yz}{1-z^2}\frac{\partial }{\partial y} + \frac{\partial }{\partial z}. \]

So then some much more unpleasant calculations using (1.3) shows you that \([X,Y] = 0\) in these coordinates as well.

Now we look at some examples of Lie algebras not coming from vector fields.

More generally, the Jacobi identity records the failure of associativity of the Lie bracket:

1.7 Integral Curves and Lie Derivatives

To motivate the definition of the Lie bracket, I claimed that there was additional algebraic structure on \(\mathfrak {X}(M)\) in the form of a binary operation, and then (hopefully!) convinced you that there was really only one sensible way to define such an operation.

But it’s probably not at all obvious why one should have guessed that there was a binary operation on vector fields in the first place, and, though I don’t actually know the history, my guess is that this was not the original motivation for the Lie bracket.

In some sense the more basic notion is that of the Lie derivative, which is a sort of directional derivative of vector fields which, as we will see, actually generalizes to arbitrary tensor fields.

The idea is that, given two vector fields \(X\) and \(Y\) on a manifold \(M\), we might like to define a “directional derivative of \(Y\) in the direction of \(X\)” operator, which we will denote \(\mathcal{L}_XY\), where the \(\mathcal{L}\) is for “Lie derivative.” It will turn out that \(\mathcal{L}_XY = [X,Y]\), which might be surprising, but if you look back to the local coordinate expression (1.3) for \([X,Y]\), you’ll notice it involved differentiating the coefficients of \(Y\) with respect to \(X\) (and also \(X\) with respect to \(Y\), but we’ll shortly see why there’s this symmetry).

It’s worth thinking for yourself about how you might go about defining such a derivative operator, so I would encourage you to do so before reading on.

Here’s a way of making this “directional derivative for vector fields” operation more precise: say we have our vector fields \(X\) and \(Y\) and we want to compute \(\mathcal{L}_XY\) at a point \(p \in M\). Presumably the idea is to measure how much \(Y\) is changing as we vary \(p\) in the direction of \(X\).

In other words, we want to integrate \(X\) to get a curve \(\alpha : (-\epsilon , \epsilon ) \to M\) so that \(\alpha (0) = p\) and \(\alpha '(t) = X(\alpha (t))\) for all \(t \in (-\epsilon , \epsilon )\), then look at something like

See Figure 1.12.

In other words, we look at the rate of change of \(Y\) as we flow in the direction of \(X\). Unfortunately, as written the above difference quotient doesn’t make sense, but first let’s talk about flows and integrating vector fields.

We often use the notation \(\Phi _t(q):=\Phi (t,q)\) and call \(\Phi _t\) the local flow of \(X\).

Example 1.7.3

Let \(M = S^2\) and consider the vector field \(X(x,y,z) = zx \frac{\partial }{\partial x} + zy \frac{\partial }{\partial y} +\left(z^2-1\right) \frac{\partial }{\partial z}\) (of course, this is written in extrinsic \(\mathbb {R}^3\) coordinates rather than intrinsic coordinates). See Figure 1.14.

In fact, you can check that this is just \((d\phi _N)Y\), where \(Y\) is the vector field on \(\mathbb {R}^2\) given by \(Y(u,v) = -u \frac{\partial }{\partial u} - v \frac{\partial }{\partial v}\).

Visually, the local flow pushes the mass of the sphere down towards the south pole. We can integrate the flow explicitly in Mathematica to get

\[ \Phi _t(x,y,z) = \frac{1}{1+2e^{2t} + \left(1-e^{2t}\right)z}\left(2e^t x, 2e^t y, 1-e^{2t} + \left(1+e^{2t}\right)z\right). \]

See Figure 1.15.

This is an example of a complete vector field since the local flow exists for all time at all points. If we had removed the Antarctic Circle from the sphere, this vector field would be incomplete (since the flow ceases to exist once you go over the edge).

With local flows in our toolkit we would then like to define the Lie derivative as something like

But if you think about it, the numerator in the difference quotient makes no sense! After all, \(Y(\Phi _t(p)) \in T_{\Phi _t(p)}M\) and \(Y(p) \in T_pM\) live in completely different vector spaces. There’s no sense in which we can subtract these two vectors.

So we need some way to get the tangent vectors determined by \(Y\) at different points to be in the same tangent space.

Thus far, the only machines we have for moving tangent vectors between different tangent spaces are differentials of smooth maps. The only smooth map we have at our disposal is \(\Phi _t\), so somehow that must come into play. One possibility is

since both \(Y(\Phi _t(p))\) and \((d \Phi _t)_pY(p)\) live in \(T_{\Phi _t(p)}M\). The problem with this is that, as \(t\) changes, \(T_{\Phi _t(p)}M\) is also changing, so we’re taking limits of vectors in different tangent spaces. Now, all these tangent vectors still live in \(TM\), so this is in fact doable, but it would be better if all the vectors in the limit actually lived in the tangent space, ideally \(T_pM\).

We are finally ready to define the Lie derivative:

This is a lot of obnoxious notation, but let’s try to unpack it. First of all, notice that \(\Phi _{-t}(\Phi _t(p)) = p\), since we’re just flowing forward in time by \(t\) and then backwards for the same time. So

Hence, plugging in \(Y(\Phi _t(p))\) gives \((d\Phi _{-t})_{\Phi _t(p)}Y(\Phi _t(p)) \in T_pM\), and the difference quotient and the limit make sense as happening entirely in \(T_pM\). A (somewhat messy) picture is shown in Figure 1.16.

In looking at Figure 1.16, recall that the point was that we wanted to see how \(Y\) varied as we moved in the direction of \(X\). So at a point \(p \in M\), we follow the local flow of \(X\) for a small time \(t\). \(Y\) determines a tangent vector at the resulting point \(\Phi _t(p)\). Now we push this vector \(Y(\Phi _t(p))\) to \(T_pM\) by the negative local flow of \(X\) and compare to \(Y(p)\).

You can imagine that if \(Y\) had a constant magnitude and made a constant angle with \(X\), then pushing \(Y\) forward by the negative local flow (which is exactly \((d\Phi _{-t})_{\Phi _t(p)}Y(\Phi _t(p))\)) would just give you the same vector you started with, namely \(Y(p)\). But this makes sense: if \(Y\) has a constant magnitude and makes a constant angle with \(X\), then it’s not changing at all with respect to \(X\), and this derivative should really be 0. ¹¹

The slightly amazing fact is that the Lie derivative and the Lie bracket are the same:

Notice, in particular, that this shows that the Lie derivative is antisymmetric:

by the antisymmetry of the Lie bracket, which you might not have guessed from the definition of the Lie derivative.

Let’s focus on the first term in the numerator, \(\left((d\Phi _{-t})_{\Phi _t(p)}Y)f\right)(p)\):

where we used Lemma 1.3.4 for the first and last equalities ¹² and the chain rule for the middle equality. Therefore,

If we could turn this into something like

we’d be in business.

In fact, if we could write \(f(\Phi _{-t}(q)) = f(q) + t g(t,q)\) for some function \(g\), then we could get a term like (1.5) to appear because then we’d have

This will then equal \(((XY-YX)f)(p)\) if we can find such a \(g\) so that the expression \(Y(g(0,p))\) is equal to \((Y(Xf))(p)\).

But this is easy: rearranging \(f(\Phi _{-t}(q)) = f(q) + tg(t,q)\) yields

at least for \(t \neq 0\). We can extend the definition to \(t=0\) by taking the limit:

which is exactly what we need, since then \(Y(g(0,p)) = (Y(Xf))(p)\).

1.8 Extended Example on \(S^3\)

Let \(S^3\) be the unit sphere in \(\mathbb {R}^4 \cong \mathbb {C}^2 \cong \mathbb {H}\), where

is the division ring of quaternions. We have the defining conditions

which for example imply that

and

so \(\mathbb {H}\) is noncommutative. \(\mathbb {H}\) also has a conjugation operation defined by

and an inner product given by,

When \(p = a+bi+cj+dk\) and \(q = t + xi + yj + zk\), we see that

so this agrees with the usual dot product on \(\mathbb {R}^4\). Moreover, the induced norm

is the standard one.

Thought of as the unit quaternions \(\{ q \in \mathbb {H}: \| q\| = 1\} \), it becomes clear that \(S^3\) is a group: if \(p = a+bi+cj+dk , q = t + xi + yj + zk \in \mathbb {H}\) are both unit quaternions, meaning that

then their product is also a unit quaternion:

To give some other terminology, \(S^3\) is the symplectic group \(\operatorname {Sp}(1)\); that is, the quaternionic analog of the unitary group for \(\mathbb {H}^1\). (We will eventually see that this group is also isomorphic to \(\operatorname {SU}(2)\) and \(\operatorname {Spin}(3)\)).

Geometrically, it is clear that, for any \(p \in S^3\), the tangent space \(T_pS^3\) can be thought of as the quaternions which are orthogonal to \(p\). So for example the tangent space to the identity \(1 \in S^3\) is

the purely imaginary quaternions. Using the same idea as in Example 1.5.3, we can push any tangent vector at the identity around to get a left-invariant vector field on all of \(S^3\). In particular, we get three mutually perpendicular vector fields \(X,Y,Z \in \mathfrak {X}(S^3)\) given by

for all \(p \in S^3\). (Strictly speaking, \(X(p) = (d L_p)_1 i\), where \(L_p\) is left-multiplication by \(p\), and \(i\) is intepreted as an element of \(T_1S^3\) [and similarly for \(Y\) and \(Z\)], but \(L_p\) being linear implies that \((dL_p)_1 i = pi\).)

To check that, say, \(X(p)\) is really a tangent vector at \(p\), notice that

so \(X(p) \bot p\) and hence \(X(p) \in T_pS^3\). Similar calculations show that \(Y(p),Z(p) \in T_p S^3\).

Moreover \(X\), \(Y\), and \(Z\) are mutually perpendicular everywhere: for example

(A better argument: multiplication by a unit quaternion is an isometry of \(\mathbb {H}= \mathbb {R}^4\), so orthogonality of \(X(1) = i\) and \(Y(1) = j\) carries over to \(T_pS^3\).)

Now, the corresponding local flows are

as we can see by differentiating; for example,

Now we compute some Lie brackets. For example,

The differential \((d\xi _{-t})_{\xi _t(p)}\colon \thinspace \! T_{\xi _t(p)}S^3 \to T_p S^3\), but we can interpret both of these as subspaces of the ambient space \(\mathbb {R}^4\), so the differential can be represented by a \(4 \times 4\) matrix, namely

so

Putting this all together, then

In other words, \([X,Y] = 2Z\) and, by similar arguments, \([Y,Z] = 2X\) and \([Z,X] = 2Y\).

In our language from Example 1.5.3, \(X\), \(Y\), and \(Z\) are left-invariant vector fields on the Lie group \(S^3\), and so you should expect that the Lie bracket on left-invariant vector fields on \(S^3\) corresponds to a Lie algebra structure on \(T_1 S^3\), which is a 3-dimensional vector space. We’ll call this 3-dimensional Lie algebra \(\mathfrak {sp}(1)\) since \(S^3 = \operatorname {Sp}(1)\).

Thinking of \(S^3\) and \(\mathfrak {sp}(1)\) as living in the space of \(1 \times 1\) quaternionic matrices (that is, inside \(\mathbb {H}\)), we might expect that the matrix commutator on \(\mathfrak {sp}(1) = \{ xi + yj + zk\} \) should correspond to the above Lie bracket. Indeed, the commutator of the \(1 \times 1\) matrices \(i\) and \(j\) is

which agrees with the above calculation that \([X,Y] = 2Z\).

Since the only Lie algebra structure we’ve seen on a 3-dimensional vector space is that of \((\mathbb {R}^3,\times )\), or equivalently \((\mathfrak {so}(3),[\cdot ,\cdot ])\), you might guess that \(\mathfrak {sp}(1)\) is just another iteration of the same Lie algebra.

Indeed, if \(e_1, e_2 , e_3\) is the standard basis for \(\mathbb {R}^3\) and we define \(F\colon \thinspace \! \mathbb {R}^3 \to \mathfrak {sp}(1)\) by

then we see that, e.g.,

and more generally it’s easy to check that \(F(u\times v) = [F(u),F(v)]\) for any \(u,v \in \mathbb {R}^3\), so \(F\) is an isomorphism of Lie algebras.

This also implies that \(\mathfrak {so}(3)\) and \(\mathfrak {sp}(1)\) are isomorphic Lie algebras, even though the Lie groups \(\operatorname {SO}(3)\) and \(S^3\) are not even homotopy equivalent (for example, \(\pi _1(\operatorname {SO}(3))\cong \mathbb {Z}/2\mathbb {Z}\) and \(\pi _1(S^3) = \{ 1\} \)), which is going to imply that they cannot be isomorphic as Lie groups. ¹³

2.1 Integration, Forms, and an Informal Definition

Now that we’ve defined vector fields and explored some special features (like the Lie bracket), the next objects to turn our attention to are differential forms. Differential forms turn out to be the right tool with which to define integration on manifolds, and they also turn out to encode the cohomology ring of a manifold, which means they have both a natural binary operation (corresponding to the cup product in cohomology) and a natural derivative operation (corresponding to the coboundary map).

To try to build up some intuition for differential forms, think back to vector calculus and computing integrals of the form

where \(A \subseteq \mathbb {R}^n\) is an open set (or maybe the closure of an open set), and \(f: A \to \mathbb {R}\) is a (sufficiently nice) function. ¹⁴ If you stare at (2.1), and in particular at the expression \(f\, dx_1 \dots dx_n\) being integrated, what exactly is this thing?

To get a handle on what it is, it’s helpful to see how it transforms. If \(B \subseteq \mathbb {R}^n\) is open and \(g: B \to A\) is a diffeomorphism, ¹⁵ then

where I’m using \(y_1, \dots , y_n\) to indicate the coordinates on \(B\) and \(Jg\) is the Jacobian matrix of \(g\). So now whatever kind of object \(f\, dx_1 \dots dx_n\) is, it’s the same type of object as \((f \circ g) \, |\! \det Jg|\, dy_1 \dots dy_n\), and this describes how it transforms under diffeomorphisms.

Now, examining \((f \circ g)\, |\! \det Jg|\, dy_1 \dots dy_n\), it’s clear that this is a more complicated object than just a function. In particular, the presence of the determinant tells you that this is some sort of alternating, multilinear object. Recall the definitions:

Thinking of the determinant map as a function \(\det \colon \thinspace \! V^n \to \mathbb {R}\) where \((v_1, \dots , v_n) \in V^n\) are interpreted as the columns of a matrix whose determinant is then computed, \(\det \) is multilinear and alternating. Indeed, these two properties uniquely characterize the determinant up to scale:

The point of all of this is that the things we actually integrate in multivariable calculus are alternating, multilinear gadgets. Informally, this is how we should think about differential forms on manifolds:

In other words, at each point \(p \in M\), a \(k\)-form \(\omega \) will input \(k\) tangent vectors at \(p\) and output a real number in an alternating, multilinear way. You might guess that this means that \(\omega \) is really a (smooth) section of some vector bundle over \(M\), and indeed we will shortly define it in this way.

But before that, let’s look at some examples of things that should be differential forms according to whatever formal definition we eventually give.

2.2 Differential Forms and Vector Calculus

Suppose we restrict to the case of a (Riemannian) 3-manifold; that is, a 3-dimensional manifold \(M\) endowed with a Riemannian metric \(g\) (again, a Riemannian metric is a [smooth] choice of inner product \(g_p\) on each tangent space; if this makes you uncomfortable, just assume \(M = \mathbb {R}^3\) and we just have the standard dot product on each tangent space). This section is a bit weird and definitely very informal: I’m using a bunch of stuff that I haven’t actually defined yet. But the point is to try to say that concepts like differential forms and exterior derivatives that we will eventually define carefully are just generalizations of things you already understand very well.

From ex:0-forms and functionsex:n-1 forms, we know that

So anything we can say about differential forms on \(M\) must be expressible just in terms of vector fields and functions. We know from vector calculus that there are various differentiation operators on functions and vector fields, namely div, grad, and curl, so what do these mean at the level of forms.

Or, turning it around, are there operations on forms which fill in the squares in the following diagram and makes them commute?

First, we (hopefully) recall from vector calculus that the gradient of a function is a vector field, which can be computed in local coordinates as

Then the corresponding 1-form \((\nabla f)^\flat \) is defined at \(p \in M\) by

where \(Y = \sum _{i=1}^3 b_i \frac{\partial }{\partial x_i}\) in local coordinates and I’ve written \(f_{x_i}\) as a shorthand for \(\frac{\partial f}{\partial x_i}\).

If we were able to arrange it so that the local coordinate basis \(\left\{ \frac{\partial }{\partial x_1},\frac{\partial }{\partial x_2},\frac{\partial }{\partial x_3}\right\} \) were orthonormal with respect to \(g_p\), then this would simplify as

by Lemma 1.3.4. In other words, \((\nabla f)^\flat = df\).

Recall (Example 2.1.8) that at each point a 1-form corresponds to a cotangent vector, so the local coordinate basis \(\left\{ \frac{\partial }{\partial x_1},\frac{\partial }{\partial x_2},\frac{\partial }{\partial x_3}\right\} \) induces a dual basis which we call \(\{ dx_1, dx_2, dx_3\} \) for \(\left(T_pM\right)^\ast \) defined by

In these terms,

In other words, in the presence of a Riemannian metric, ¹⁹ computing the differential of a function on a 3-manifold (in fact, this works equally well on an \(n\)-manifold) corresponds to taking the gradient of the function.

Turning to curl, recall that, if \(X \in \mathfrak {X}(M)\) is written in local coordinates as \(X = a_1 \frac{\partial }{\partial x_1} + a_2 \frac{\partial }{\partial x_2} + a_3 \frac{\partial }{\partial x_3}\), then

As discussed in Example 2.1.12, this corresponds to an element of \(\omega ^{3-1}(M) = \Omega ^2(M)\) given by

For reasons to be explained later, I’m going to call this 2-form \(\star (\nabla \times X)^\flat \).

So now what is the operation on \(X^\flat = a_1 dx_1 + a_2 dx_2 + a_3 dx_3\) which would have produced \(\star (\nabla \times X)^\flat \)? The idea is to compute the differential (which is a 1-form) of each coefficient function, and combine it with the corresponding \(dx_i\) in a particular way:

where the rule is that \(dx_i \wedge dx_j = - dx_j \wedge dx_i\), which is just the alternating condition on forms. So, if you buy the above manipulations on some level, there is a fairly straightforward generalization of the differential which corresponds exactly to the curl in this setting.

Finally, if \(X \in \mathfrak {X}(M)\) is written in local coordinates as \(X = a_1 \frac{\partial }{\partial x_1} + a_2 \frac{\partial }{\partial x_2} + a_3 \frac{\partial }{\partial x_3}\), then the divergence is

which is a function. But the isomorphism \(C^\infty (M) \cong \Omega ^3(M)\) described in Example 2.1.11 says that this function corresponds to some 3-form

because for an appropriate choice of coordinates I can write \(\operatorname {dVol}_M = dx_1 \wedge dx_2 \wedge dx_3\).

Can we get to this 3-form by applying our generalized differential to the 2-form \(\star X^\flat = a_1 dx_2 \wedge dx_3 + a_2 dx_3 \wedge dx_1 + a_3 dx_1 \wedge dx_2\)? Yes!

The upshot is that we’ve now filled in the diagram:

Of course, I haven’t really given a rigorous definition of \(d\) (which is called the exterior derivative), nor of \(\wedge \) (the wedge product), but, as we will see, there is a natural, coordinate-free way of defining these things for arbitrary differential forms on arbitrary manifolds which specializes to the calculuations above in local coordinates (and hence corresponds to gradient and divergence on manifolds of arbitrary dimension).

Hopefully the fact that div, grad, and curl are all essentially the same operator in this framework helps convince you that differential forms are useful and important. In fact, it gets even better: in this language, the Fundamental Theorem of Calculus, Green’s Theorem, and the Divergence Theorem all turn out to be special cases of the same theorem: Stokes’ Theorem for differential forms.

More than just generalizing essentially all of vector calculus, differential forms also encode topological information (in the form of cohomology). To see this, look again at the diagram above:

1. We know from vector calculus that \(\nabla \times (\nabla f) = 0\) for any \(f \in C^\infty (M)\) and \(\nabla \cdot (\nabla \times X) = 0\) for any \(X \in \mathfrak {X}(M)\). Since the diagram commutes, this implies that \(d \circ d = 0\) regardless of whether you start in \(\Omega ^0(M)\) or \(\Omega ^1(M)\) (and in fact this is still true if you start in \(\Omega ^2(M)\) or \(\Omega ^3(M)\) since \(\Omega ^4(M) = \{ 0\} \)). In other words,

   \[ \Omega ^0(M) \longrightarrow \Omega ^1(M) \longrightarrow \Omega ^2(M) \longrightarrow \Omega ^3(M) \]

   is a (co)chain complex. More generally, if we add indices to the exterior derivatives to indicate where they start and end,

   \[ \Omega ^0(M) \stackrel{d_0}{\longrightarrow } \Omega ^1(M) \stackrel{d_1}{\longrightarrow } \dots \stackrel{d_{n-2}}{\longrightarrow } \Omega ^{n-1}(M) \stackrel{d_{n-1}}{\longrightarrow } \Omega ^n(M) \]

   is always a cochain complex on any \(n\)-manifold (i.e., \(d_k \circ d_{k-1} = 0\) for all \(k\)). The usual thing you do when you have a (co)chain complex is to take the (co)homology, and that’s also useful in this case. Doing so yields the de Rham cohomology groups

   \[ H_{\text{dR}}^k(M) := \frac{\ker d_k}{\operatorname {im} d_{k-1}}. \]

   In fact, as usual with cohomology, the cohomology groups fit together into a graded ring, and the product operation corresponds to the wedge product \(\wedge \). And de Rham cohomology will turn out to be isomorphic to singular (or simplicial) cohomology with real coefficients.

2. In our 3-manifold example, we saw that \(\Omega ^0(M) = C^\infty (M) \cong \Omega ^3(M)\) and \(\Omega ^1(M) \cong \mathfrak {X}(M) \cong \Omega ^2(M)\). In general, if \(M\) is an \(n\)-dimensional manifold, then it will turn out that \(\Omega ^k(M) \cong \Omega ^{n-k}(M)\) for all \(k=0, \dots , n\). Even better, this isomorphism descends to the de Rham cohomology groups and we will have that

   \[ H_{\text{dR}}^k(M) \cong H_{\text{dR}}^{n-k}(M) \]

   for all \(k\), which is the appropriate version of Poincaré duality for de Rham cohomology.

2.3 Tensor Algebras and Tensor Fields

In order to give a more rigorous definition of differential forms (and in particular to get them to form a graded algebra), as well as more general tensor fields, we need to do some (multilinear) algebra and talk about tensor algebras and exterior algebras.

Before diving in, let me just give you my perspective on tensors. Basically, the point is that tensors are the right tool for turning multilinear algebra into linear algebra. More precisely, suppose we have three vector spaces \(U\), \(V\), and \(W\), and a map \(F: U \times V \to W\) which is multilinear (or, really, bilinear in this case). Again, this just means that \(F\) is linear in each factor:

Returning to the general setting, we have vector space \(U\), \(V\), and \(W\) and a multilinear map \(F: U \times V \to W\). Now suppose \(w_0 \in W\) and we want to solve a problem of the form \(F(u,v) = w_0\). Since \(F\) is multilinear, you might hope that this is somehow just a linear algebra problem, which would be solvable, at least in principle. But, at least as stated, this is very much not a linear problem…

To reiterate, for me ²⁰ the point of tensor products is to turn multilinear maps and problems into linear maps and problems.

Here’s the key theorem (or, if you start form a different perspective, the below theorem is the definition):

Theorem 2.3.7 (Universal Property of the Tensor Product)

If \(\phi \colon \thinspace \! U \times V \to U \otimes V\) is the map given by \((u,v) \mapsto u \otimes v\) and \(F \colon \thinspace \! U \times V \to W\) is bilinear, then there exists a unique linear map \(\widetilde{F} \colon \thinspace \! U \otimes V \to W\) making the following diagram commute:

Moreover, this uniquely characterizes \(U \otimes V\): any vector space satisfying this property must be isomorphic to \(U \otimes V\).

Clearly, we’ve paid a price for turning bilinear maps into linear maps: whereas \(\dim (U \times V) = m+n\), we have \(\dim (U \otimes V) = mn\), which is usually a lot bigger. So we turn multilinear algebra into linear algebra at the cost of having to work in much higher dimensions.

In general, \(\mathcal{T}(V)\) is noncommutative, associative, and (bi-)graded.

Example 2.3.22

Consider the Riemmannian metric \(g = \frac{1}{y^2}(dx \otimes dx + dy \otimes dy)\) (also commonly written as \(ds^2 = \frac{1}{y^2}(dx^2 + dy^2)\)) on \(H = \{ (x,y) \in \mathbb {R}^2 : y {\gt} 0\} \). ²¹ This is the Poincaré half-plane model of the hyperbolic plane. In general, we can compute the length of a curve \(\gamma (t) = (\gamma _1(t),\gamma _2(t))\) by integrating the length of \(\gamma '(t)\) along the curve. Since \(\gamma '(t) \in T_{\gamma (t)}H\), its length is

\[ \sqrt{g_{\gamma (t)}(\gamma '(t),\gamma '(t))} = \sqrt{\frac{\gamma '(t) \cdot \gamma '(t)}{\gamma _2(t)^2}} = \frac{\| \gamma '(t)\| _E}{\gamma _2(t)}, \]

where \(\cdot \) is the standard dot product on the plane and \(\| \, \| _E\) is the Euclidean norm (I can just use \(\gamma _2(t)\) in the denominator, rather than \(|\gamma _2(t)|\), because the second coordinate is always positive in \(H\)).

It will turn out that the geodesics in this metric are the vertical lines and semicircles centered at points on the \(x\)-axis; see Figure 2.1. Let’s compute distances between points on these geodesics.

For the vertical lines, we can parametrize the line through \((x_0,0)\) by \(\gamma (t) = (x_0,t)\). So then \(\gamma '(t) = (0,1)\) and the distance between \((x_0,y_0)\) and \((x_0,y_1)\) is

\[ \int _{y_0}^{y_1} \sqrt{g_{\gamma (t)}(\gamma '(t),\gamma '(t))}\, dt = \int _{y_0}^{y_1} \frac{1}{t}\, dt = \ln {y_1} - \ln {y_0} = \ln \left(\frac{y_1}{y_0}\right). \]

Notice, in particular, that the \(x\)-axis is infinitely far away from any point in \(H\).

On the other hand, the semicircle of radius \(r\) centered at \((x_0,0)\) can be parametrized by \(\gamma (t) = (x_0 + r \cos t, r \sin t)\). Now \(\gamma '(t) = r(-\sin t, \cos t)\), so the distance between \(\gamma (\theta _0)\) and \(\gamma (\theta _1)\) is

$$\begin{multline*} \int _{\theta _0}^{\theta _1} \sqrt{g_{\gamma (t)}(\gamma '(t),\gamma '(t))}\, dt = \int _{\theta _0}^{\theta _1} \frac{r}{r \sin t} \, dt = \int _{\theta _0}^{\theta _1} \csc t\, dt = \operatorname {arctanh}(\cos \theta _0) - \operatorname {arctanh}(\cos \theta _1). \end{multline*}$$

2.4 Exterior Algebras and Differential Forms

We just saw that 1-forms are \((0,1)\)-tensor fields, but in general it’s not quite natural to think of \(k\)-forms as tensor fields because we need to incorporate the alternating condition. We could do this by considering the subset of alternating tensor fields, but it turns out to be more natural to take a quotient of the tensor algebra to get something called the exterior algebra.

Specifically, just as the tensor product is a way of turning multilinear maps into linear maps, the exterior product will turn out to be the way to turn alternating multilinear maps into linear maps. Even though we haven’t actually defined the exterior product yet, it might seem silly to introduce another new product; after all, alternating multilinear maps are multilinear so the tensor product already turns them into linear maps.

That being the case, the exterior product does this in much lower dimension, so it’s still worthwhile. At the most extreme level, you can see this because the natural space in which the tensor product lives (as a binary operation) is the tensor algebra \(\mathcal{T}(V)\) (or, if you like, only the \((r,0)\) part \(V_0^0 \otimes V_1^0 \otimes V_2^0 \otimes \dots \)), which is always infinite-dimensional. Whereas the exterior product will be a binary operation on the exterior algebra, which will turn out to be finite-dimensional if \(V\) is. And, if you weren’t already aware, finite-dimensional vector spaces are much, much nicer than infinite-dimensional vector spaces: it’s always worthwhile to make the effort to work in finite dimensions if at all possible.

The grading is given by the degree of the wedge product, since

where \({\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }(V) = V_k^0/\mathcal{I}_k\), and \(\mathcal{I}_k = \mathcal{I}(V) \cap V_k^0\).

In particular, this says that we can always rearrange any wedge product to have the terms in any order we like. For example, if the terms are indexed (e.g., if they come from some ordered basis), we can always make the indices appear in increasing order at the cost of a sign.

This is exactly the finite-dimensionality described above. Even if we’re just interested in alternating, bilinear maps \(V \times V \to W\), \(\dim {\def\tmp {2} {\textstyle \bigwedge \! \! ^{2}} }(V) = \binom {n}{2}\) is less than half of \(\dim V^{\otimes 2} = n^2\), so working with the exterior power rather than the tensor power means we’re doing linear algebra in half the dimensions, which is almost always worthwhile computationally.

As you would expect, the exterior products satisfy a universal property analogous to the one satisfied by tensor products:

Theorem 2.4.7 (Universal Property of the Exterior Product)

If \(\phi \colon \thinspace \! V^k \to {\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }(V)\) is the map given by \((v_1, \dots , v_k) \mapsto v_1 \wedge \dots \wedge v_k\) and \(F \colon \thinspace \! V^k \to W\) is an alternating multilinear map, then there exists a unique linear map \(\widetilde{F} \colon \thinspace \! {\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }(V) \to W\) making the following diagram commute:

Again, this uniquely characterizes \({\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }(V)\) up to isomorphism.

If we let \(W = \mathbb {R}\) (or whatever the base field is) in Theorem 2.4.7, then we see that alternating multilinear functionals on \(V^k\) correspond precisely to linear functionals on \({\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }V\); that is, to elements of the dual space \(\left({\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }V\right)^\ast \). But of course this is essentially how we “defined” differential \(k\)-forms in Definition 2.1.4: at each point a \(k\)-form should exactly be an alternating multilinear functional on the product of \(k\) copies of the tangent space at that point.

If you’ve been reading very closely, you might have noticed that I played fast and loose with parentheses and stars above. The discussion in the paragraph before Definition 2.4.9 suggests that, if \(\omega \) is a \(k\)-form (that is, an alternating, multilinear map \(\mathfrak {X}(M)^k \to C^\infty (M)\) according to our informal definition), then at a point \(p \in M\) we should expect that \(\omega (p) \in \left({\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }\left(T_pM\right)\right)^\ast \). But I’ve just defined \(\omega (p)\) as an element of \({\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }\left(\left(T_pM\right)^\ast \right)\)! So what’s going on?

First of all, \({\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }\left(\left(T_pM\right)^\ast \right)\) is (at least to me) conceptually nicer than \(\left({\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }\left(T_pM\right)\right)^\ast \): it’s an exterior power itself, rather than being the dual of an exterior power, and so it obviously fits into an exterior algebra, namely \({\def\tmp {} {\textstyle \bigwedge \! \! ^{}} }\left(\left(T_pM\right)^\ast \right)\). So that’s the reason for wanting to define exterior bundles in terms of \({\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }\left(\left(T_pM\right)^\ast \right)\) rather than \(\left({\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }\left(T_pM\right)\right)^\ast \).

Of course, the reason I can do this is because they’re isomorphic:

The strategy here is to find a nondegenerate pairing between \({\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }(V^\ast )\) and \({\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} } (V)\). Recall that a nondegenerate pairing of vector space \(V\) and \(W\) is a bilinear map

with the property that, for any nonzero \(w \in W\), there exists \(v \in V\) so that \((v,w) \neq 0\) (and similarly when you swap the roles of \(v\) and \(w\)). Notice, in particular, that a pairing corresponds uniquely to a linear map \(V \otimes W \to \mathbb {R}\); that is, an element of \((V \otimes W)^\ast \).

If there is a nondegenerate pairing of \(V\) and \(W\), then it induces injective maps \(\phi \colon \thinspace \! V \to W^\ast \) and \(\psi \colon \thinspace \! W \to V^\ast \) given by

In finite dimensions, we know that \(V \cong V^\ast \) and \(W \cong W^\ast \), so the composition \(V \overset {\phi }{\hookrightarrow } W^\ast \cong W \overset {\psi }{\hookrightarrow } V^\ast \) gives an injective (and therefore also surjective) map \(V \to V^\ast \), which implies that \(\phi \) and \(\psi \) had to be bijective, and therefore isomorphisms. In particular, this implies that \(V \cong W^\ast \) (and therefore also \(W \cong V^\ast \) and indeed \(V \cong W\)).

So the strategy for proving Lemma 2.4.10 is to find a nondegenerate pairing of \({\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }(V^\ast )\) and \({\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }(V)\), which will imply the result. ²²

Since arbitrary elements of \({\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }(V^\ast )\) and \({\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }(V)\) are linear combinations of decomposable elements, we define \((\cdot , \cdot ) \colon \thinspace \! {\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }(V^\ast ) \times {\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} } (V) \to \mathbb {R}\) by extending the above linearly.

It is straightforward (but somewhat tedious) to check that \((\cdot , \cdot )\) is nondegenerate, and thus induces the desired isomorphism \({\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }(V^\ast ) \overset {\cong }{\to } \left({\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }(V)\right)^\ast \). Doing this for each \(k\) shows that the exterior algebra

Notice that this argument—and especially (2.4)—really validates the idea that all alternating multilinear maps are, in some sense, determinants.

2.5 The Exterior Derivative

We’ve already seen some calculations of the exterior derivative in local coordinates in Section 2.2, but to give a coordinate-free definition of the exterior derivative we need to talk about derivations.

The idea is that we kind of know what such an thing should look like in local coordinates around a point \(p \in M\): any \(k\)-form \(\omega \in \Omega ^k(M)\) can be written in local coordinates as

So then define

I claim that (2.5) is the only way to get an anti-derivation of degree \(+1\) that satisfies 1 and 2 in local coordinates. Why? If \(d\) satisfies 1 and 2, then \(dx_i = d(x_i)\) is the result of applying \(d\) to the 0-form \(x_i\) (by 1), so \(d(dx_i) = (d \circ d)(x_i) = 0\) (by 2). But then the anti-derivation property implies that

since each \(d(dx_{i_j}) = 0\). But then, again applying the anti-derivation condition together with linearity,

(a)

This is an anti-derivation.

(b)

It satisfies the cocycle condition 2.

(c)

Uniqueness.

(a) and (b) are both computations. Here’s the computation for (b): If \(\omega = \sum _I f_I\, dx_I\) in a neighborhood of \(p \in M\), then

in that same neighborhood, so

since \(\frac{\partial ^2 f}{\partial x_j\partial x_i} = \frac{\partial ^2 f}{\partial x_i\partial x_j}\) but \(dx_j \wedge dx_i = - dx_i \wedge dx_j\).

Turning to (c), suppose there exists a degree \(+1\) anti-derivation \(D\colon \thinspace \! \Omega ^\ast (M) \to \Omega ^\ast (M)\) satisfying 1 and 2. In particular, \(Df = df\) for all \(f \in C^\infty (M) = \Omega ^0(M)\). Then

since \(\displaystyle D(dx_{i_j}) \mathop{=}\limits ^{\ref{it:d_0 is differential}} D(Dx_{i_j}) \mathop{=}\limits ^{\ref{it:d^2=0}} 0\) by hypothesis.

Therefore, if \(\omega = \sum _I f_I dx_I\) in a neighborhood of \(p\), by linearity and the anti-derivation condition we have

in that neighborhood. Since this is true for all \(p \in M\), it must be the case that \(D\omega = d\omega \).

Notice that \(d\) being degree \(+1\) and satisfying the cocycle condition 2 says exactly that we have a diagram

which is a cochain complex.

2.6 Pullbacks

Suppose we have manifolds \(M\) and \(N\) and a smooth map \(f \colon \thinspace \! M \to N\). Since we have a change of variables formula for integrals, and differential forms are supposed to be the objects that we can integrate on manifolds, we should expect that \(f\) gives a way of turning differential forms on one of \(M\) or \(N\) into forms on the other. If you think about how the change of variables formula works, you can figure out which direction this is supposed to go, but let’s build up to this a bit more slowly.

First of all, recall that if \(v \in T_pM\), then \(df_p v \in T_{f(p)}N\); that is, \(df_p \colon \thinspace \! T_pM \to T_{f(p)}N\). More precisely, since \(v = \alpha '(0)\) for some smooth curve \(\alpha \) with \(\alpha (0) = p\), we defined \(df_pv = (f \circ \alpha )'(0)\). In words, \(f\) gives a way of pushing tangent vectors forward using \(df\), ²³ and we could have guessed this in advance because \(v\) is defined in terms of a map into \(M\), which can then be composed with \(f\) to get a map into \(N\). In categorical terms, tangent vectors and vector fields are transformed covariantly, meaning that tangent vectors or vector fields get sent in the same direction (from \(M\) to \(N\)) as the original map \(f\colon \thinspace \! M \to N\).

Now let’s think about how we could move differential forms around using \(f\). At a point, a \(k\)-form \(\omega \in \Omega ^k(M)\) corresponds to \(\omega _p \in {\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }\left(\left(T_pM\right)^\ast \right) \cong \left({\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }\left(T_pM\right)\right)^\ast \), or equivalently an alternating multilinear functional \(\left(T_pM\right)^k \to \mathbb {R}\). In other words, unlike a tangent vector which is defined by a map into \(M\), a differential form corresponds to a map out of \(\left(T_pM\right)^k\). That means there’s really no way to compose \(\omega _p \in {\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} }\left(\left(T_pM\right)^\ast \right)\) with \(f\) to get something on \(N\).

However, if we start instead with \(\eta \in \Omega ^k(N)\), then, as above, at a point \(q \in N\), \(\eta _q\) corresponds to a map \(\left(T_qN\right)^k \to \mathbb {R}\), so we should be able to precompose with a map \(\left(T_pM\right)^k \to \left(T_q N \right)^k\) to get a \(k\)-form on \(M\). More precisely:

Notice, first of all, that this definition makes sense because \(df_p\colon \thinspace \! T_pM \to T_{f(p)}N\), so each \(df_p v_i \in T_{f(p)}N\), and hence can be fed into \(\eta _{f(p)}\).

In the paragraph before Definition 2.6.1 I said we could precompose \(\eta \) with a map \(\left(T_pM\right)^k \to \left(T_q N \right)^k\), but then didn’t write down such a map. What’s going on here? The differential \(df_p \colon \thinspace \! T_p M \to T_{f(p)}N\) induces a map \(\left(df_p\right)^k \colon \thinspace \! \left(T_p M\right)^k \to \left(T_{f(p)}N\right)^k\) by just applying \(df_p\) separately in each factor. So then another way of stating Definition 2.6.1 is that

Example 2.6.3

Let \(f \colon \thinspace \! S^1 \to \mathbb {R}^2\) be the inclusion of the unit circle into the plane; i.e., \(f(\theta ) = (\cos \theta , \sin \theta )\). Let \(\omega = dx \in \Omega ^1(\mathbb {R}^2)\), which at each point \(p \in \mathbb {R}^2\) is computed by \(dx\left(a \frac{\partial }{\partial x} + b \frac{\partial }{\partial y}\right) = a\); or, said another way, by taking the dot product of the input tangent vector with the first standard basis vector.

What is \(f^\ast \omega = f^\ast dx\)? At any point \(\theta _0 \in S^1\), we have the local coordinate basis \(\left\{ \frac{\partial }{\partial \theta }\right\} \) for \(T_{\theta _0}S^1\), so we can completely specify \(f^\ast dx\) by determining the value of

\[ (f^\ast dx)_{\theta _0}\left( a \frac{\partial }{\partial \theta }\right) \]

for any \(a \in \mathbb {R}\). By Definition 2.6.1,

\[ (f^\ast dx)_{\theta _0}\left( a \frac{\partial }{\partial \theta }\right) = (dx)_{f(\theta _0)}\left(df_{\theta _0}\left( a \frac{\partial }{\partial \theta }\right)\right). \]

So first we’ll need to compute \(df_{\theta _0}\left( a \frac{\partial }{\partial \theta }\right)\). We can write \(a \frac{\partial }{\partial \theta } = \alpha '(0)\), where \(\alpha (t) = \theta _0 + a t\), so Definition 1.3.1 tells us that

$$\begin{multline*} df_{\theta _0}\left( a \frac{\partial }{\partial \theta }\right) = (f \circ \alpha )'(0) = \left. \frac{d}{dt} \right|_{t=0} f(\alpha (t)) = \left. \frac{d}{dt} \right|_{t=0} f(\theta _0 + at) = \left. \frac{d}{dt} \right|_{t=0} (\cos (\theta _0 + at), \sin (\theta _0 + at)) \\ = \left. (-a\sin (\theta _0 + at),a \cos (\theta _0 + at))\right|_{t=0} = (-a \sin (\theta _0), a \cos (\theta _0)) = -a \sin (\theta _0) \frac{\partial }{\partial x} + a \cos (\theta _0) \frac{\partial }{\partial y} \end{multline*}$$

after switching between tuple notation and local coordinate basis notation.

Therefore,

\[ (f^\ast dx)_{\theta _0}\left( a \frac{\partial }{\partial \theta }\right) = (dx)_{f(\theta _0)}\left(df_{\theta _0}\left( a \frac{\partial }{\partial \theta }\right)\right) = (dx)_{f(\theta _0)}\left(-a \sin (\theta _0) \frac{\partial }{\partial x} + a \cos (\theta _0) \frac{\partial }{\partial y}\right) = -a \sin (\theta _0). \]

In other words,

\[ (f^\ast dx)_{\theta _0} = -\sin \theta _0 (d\theta )_{\theta _0} \]

or, by slight abuse of notation,

\[ f^\ast dx = -\sin \theta \, d\theta . \]

Recalling that \(\theta \in S^1\) maps to the point \((\cos \theta , \sin \theta )\), we can see that \(\sin \theta \) is the \(y\)-coordinate, so this could also be written as

\[ f^\ast dx = -y \, d\theta . \]

Geometrically, this makes sense: at the point \((x,y) = (\cos \theta , \sin \theta )\) this form is dual to the tangent vector \(-y \frac{\partial }{\partial \theta }\), which at each point is just the orthogonal projection of the first standard basis vector \(\frac{\partial }{\partial x}\) onto the tangent space to the circle, and of course \(\frac{\partial }{\partial x}\) is the tangent vector to \((x,y) \in \mathbb {R}^2\) dual to \(dx\). See Figure 2.2.

One nice thing about pullbacks is that they commute with the exterior derivative:

Before proving this, let’s see it in an example:

where we used Definition 2.6.1 for the first equality, the Chain Rule for the second, and Example 2.6.2 for the third. This completes the proof in the \(k=0\) case.

Now, suppose \(k {\gt} 0\) and that \(x_1, \dots , x_n\) are local coordinates in a neighborhood of \(f(p) \in N\), meaning that \(\omega \in \Omega ^k(N)\) has the form \(\omega _{f(p)} = \sum _I g_I dx_I\) at \(f(p)\), so that

But then

where we’ve repeatedly used both the \(k=0\) case and the fact that \(f^\ast (\alpha \wedge \beta ) = (f^\ast \alpha ) \wedge (f^\ast \beta )\) (which follows from HW #2, Problem 1).

On the other hand,

so

and we conclude that the proposition is also true for \(k{\gt}0\).

2.7 Lie Derivatives and Cartan’s Magic Formula

Recall Definition 1.7.4, the definition of the Lie derivative for vector fields:

where \(\Phi _t\) is the local flow for \(X \in \mathfrak {X}(M)\). We saw in Proposition 1.7.5 that the Lie derivative is equal to the Lie bracket \([X,Y]\).

The key idea in this derivative was to compare \(Y\) at \(\Phi _t(p)\) to \(Y\) at \(p\), and to do so we needed to move \(Y(\Phi _t(p))\) from \(T_{\Phi _t(p)}M\) to \(T_pM\), which we did with the differential of the negative flow \(d\Phi _{-t}\).

We can use the same idea to get a Lie derivative for differential forms, though we have to be careful about how we move a form \(\omega \in \Omega ^k(M)\) at a point \(\Phi _t(p)\) to \(p\): whereas vector fields push forward, differential forms pull back (as we’ve just seen in Section 2.6). So rather than pushing forward with \(d\Phi _{-t}\), we will pull back by \(\Phi _t^\ast \):

Even more generally, we can combine pushforwards and pullbacks to get Lie derivatives of arbitrary tensor fields:

Example 2.7.4

If \(M\) is a manifold and \(g\) is a Riemannian metric on \(M\), a vector field \(X \in \mathfrak {X}(M)\) is called a Killing field ²⁴ if \(\mathcal{L}_Xg = 0\). Intuitively, this means that the flow generated by \(X\) is an isometry: it preserves distances. If you read about general relativity, you will often see references to Killing fields, which encode the symmetries of spacetime.

Let’s compute \(\mathcal{L}_X g\) for some vector fields \(X\) on \(S^2\), where \(g\) is the standard Riemannian metric induced by the inner product on \(\mathbb {R}^3\). If we work in cylindrical coordinates, we have the local coordinate chart \(\phi \colon \thinspace \! (0,2\pi ) \to S^2\) given by

\[ \phi (\theta , z) = \left(\sqrt{1-z^2}\cos \theta , \sqrt{1-z^2} \sin \theta , z\right). \]

So then our local coordinate basis \(\left\{ \frac{\partial }{\partial \theta }, \frac{\partial }{\partial z} \right\} \) at a point \(p = \phi (\theta _0,z_0)\) looks like

\[ \left. \frac{\partial }{\partial \theta }\right|_p = \left. \frac{d}{dt} \right|_{t=0} \phi (\theta _0+t,z_0) = \left(-\sqrt{1-z_0^2}\sin \theta _0, \sqrt{1-z_0^2}\cos \theta _0,z_0\right) \]

and

\[ \left. \frac{\partial }{\partial z}\right|_p = \left. \frac{d}{dt} \right|_{t=0} \phi (\theta _0,z_0+t) = \left(-\frac{z_0 \cos \theta _0}{\sqrt{1-z_0^2}},-\frac{z_0 \sin \theta _0}{\sqrt{1-z_0^2}},1\right). \]

See Figure 2.3 (which is the same as Figure 1.11).

  

In turn, this implies that the Riemannian metric \(g\) is given by

$$\begin{align*} g_p\left(a \frac{\partial }{\partial \theta } + b \frac{\partial }{\partial z}, c \frac{\partial }{\partial \theta } + d \frac{\partial }{\partial z}\right) & = \left[a\left(-\sqrt{1-z_0^2}\sin \theta _0, \sqrt{1-z_0^2}\cos \theta _0,z_0\right)+ b\left(-\frac{z_0 \cos \theta _0}{\sqrt{1-z_0^2}},-\frac{z_0 \sin \theta _0}{\sqrt{1-z_0^2}},1\right)\right) \\ & \qquad \cdot \left[c\left(-\sqrt{1-z_0^2}\sin \theta _0, \sqrt{1-z_0^2}\cos \theta _0,z_0\right)+ d\left(-\frac{z_0 \cos \theta _0}{\sqrt{1-z_0^2}},-\frac{z_0 \sin \theta _0}{\sqrt{1-z_0^2}},1\right)\right] \\ & = \left(1-z_0^2\right) ac + \frac{1}{1-z_0^2} bd \end{align*}$$

or, more symbolically,

\[ g_{(\theta ,z)} = \left(1-z^2\right) d\theta \otimes d\theta + \frac{1}{1-z^2} dz \otimes dz. \]

Notice that this only depends on \(z\) and not on \(\theta \), so we would expect that flowing by \(\frac{\partial }{\partial \theta }\) should preserve this inner product; that is, \(\frac{\partial }{\partial \theta }\) should be a Killing field.

Indeed, the local flow of \(\frac{\partial }{\partial \theta }\) is given at a point \(p = \phi (\theta _0,z_0)\) by

\[ \Phi _t(p) = \phi (\theta _0+t,z) = \left(\sqrt{1-z_0^2}\cos (\theta _0+t), \sqrt{1-z_0^2} \sin (\theta _0+t). z_0\right), \]

Now, writing \(\frac{\partial }{\partial \theta } = \alpha '(0)\) where \(\alpha (s) = \phi (\theta _0+s,z_0)\), we see that

$$\begin{align*} (d\Phi _t)_p\left(\left.\frac{\partial }{\partial \theta }\right|_{p}\right) = (d\Phi _t)_p\alpha '(0) & = \left. \frac{d}{ds} \right|_{s=0} (\Phi _t \circ \alpha )(s) \\ & = \left. \frac{d}{ds} \right|_{s=0} (\Phi _t(\phi (\theta _0+s,z_0)) \\ & = \left. \frac{d}{ds} \right|_{s=0} \left(\sqrt{1-z_0^2}\cos (\theta _0+s+t), \sqrt{1-z_0^2} \sin (\theta _0+s+t), z_0\right) \\ & = \left(-\sqrt{1-z_0^2}\sin (\theta _0+t),\sqrt{1-z_0^2}\cos (\theta _0+t),0\right) \\ & = \left. \frac{\partial }{\partial \theta }\right|_{\Phi _t(p)}; \end{align*}$$

i.e., the local flow of \(\frac{\partial }{\partial \theta }\) preserves \(\frac{\partial }{\partial \theta }\), as you might expect.

On the other hand, \(\frac{\partial }{\partial z} = \beta '(0)\) where \(\beta (s) = \phi (\theta _0,z_0+s)\), so

$$\begin{align*} (d\Phi _t)_p\left(\left.\frac{\partial }{\partial z}\right|_{p}\right) = (d\Phi _t)_p\beta '(0) & = \left. \frac{d}{ds} \right|_{s=0} (\Phi _t \circ \beta )(s) \\ & = \left. \frac{d}{ds} \right|_{s=0} (\Phi _t(\phi (\theta _0,z_0+s)) \\ & = \left. \frac{d}{ds} \right|_{s=0} \left(\sqrt{1-(z_0+s)^2}\cos (\theta _0+t), \sqrt{1-(z_0+s)^2} \sin (\theta _0+t), z_0+s\right) \\ & = \left(-\frac{z_0 \cos (\theta _0+t)}{\sqrt{1-z_0^2}},-\frac{z_0 \sin (\theta _0+t)}{\sqrt{1-z_0^2}},1\right) \\ & = \left. \frac{\partial }{\partial z}\right|_{\Phi _t(p)}, \end{align*}$$

so flowing by \(\frac{\partial }{\partial \theta }\) also preserves \(\frac{\partial }{\partial z}\).

Therefore, for any \(a \left.\frac{\partial }{\partial \theta }\right|_p + b \left.\frac{\partial }{\partial z}\right|_p\in T_pS^2\), we have

\[ (d\Phi _t)_p \left(a \left.\frac{\partial }{\partial \theta }\right|_p + b \left.\frac{\partial }{\partial z}\right|_p\right) = a \left.\frac{\partial }{\partial \theta }\right|_{\Phi _t(p)} + b \left.\frac{\partial }{\partial z}\right|_{\Phi _t(p)}. \]

Since \(p\) and \(\Phi _t(p)\) have the same \(z\)-coordinate,

$$\begin{multline*} g_{\Phi _t(p)} \left( a \left.\frac{\partial }{\partial \theta }\right|_{\Phi _t(p)} + b \left.\frac{\partial }{\partial z}\right|_{\Phi _t(p)}, c \left.\frac{\partial }{\partial \theta }\right|_{\Phi _t(p)} + d \left.\frac{\partial }{\partial z}\right|_{\Phi _t(p)}\right) = (1-z_0^2)ac + \frac{1}{1-z_0^2}bd \\ = g_p\left(a \left.\frac{\partial }{\partial \theta }\right|_p + b \left.\frac{\partial }{\partial z}\right|_p,c \left.\frac{\partial }{\partial \theta }\right|_p + d \left.\frac{\partial }{\partial z}\right|_p\right). \end{multline*}$$

Hence, for any \(u,v \in T_p S^2\)

\[ (\Phi _t^\ast g_{\Phi _t(p)})(u,v) = g_{\Phi _t(p)}((d\Phi _t)_pu,(d\Phi _t)_pv) = g_p(u,v) \]

is independent of \(t\), and we see that

\[ (\mathcal{L}_{\frac{\partial }{\partial \theta }} g)_p(u,v) = \left. \frac{d}{dt}\right|_{t=0} (\Phi _t^\ast g_{\Phi _t(p)})(u,v) = 0. \]

Therefore, \(\frac{\partial }{\partial \theta }\) is a Killing field on the sphere, which makes sense: the local flow is just rotation around the \(z\)-axis, which certainly preserves distances.

On the other hand,

\[ \Psi _t(p) = \phi (\theta _0,z_0+t) = \left(\sqrt{1-(z_0+t)^2}\cos \theta _0,\sqrt{1-(z_0+t)^2}\sin \theta _0,z_0+t\right) \]

is the local flow of \(\frac{\partial }{\partial z}\). So then

$$\begin{align*} (d\Psi _t)_p\left(\left.\frac{\partial }{\partial \theta }\right|_{p}\right) = (d\Psi _t)_p\alpha '(0) & = \left. \frac{d}{ds} \right|_{s=0} (\Psi _t \circ \alpha )(s) \\ & = \left. \frac{d}{ds} \right|_{s=0} (\Psi _t(\phi (\theta _0+s,z_0)) \\ & = \left. \frac{d}{ds} \right|_{s=0} \left(\sqrt{1-(z_0+t)^2}\cos (\theta _0+s), \sqrt{1-(z_0+t)^2} \sin (\theta _0+s), z_0+t\right) \\ & = \left(-\sqrt{1-(z_0+t)^2}\sin \theta _0,\sqrt{1-(z_0+t)^2}\cos \theta _0,0\right) \\ & = \left. \frac{\partial }{\partial \theta }\right|_{\Psi _t(p)} \end{align*}$$

and

$$\begin{align*} (d\Psi _t)_p\left(\left.\frac{\partial }{\partial z}\right|_{p}\right) = (d\Psi _t)_p\beta '(0) & = \left. \frac{d}{ds} \right|_{s=0} (\Psi _t \circ \beta )(s) \\ & = \left. \frac{d}{ds} \right|_{s=0} (\Psi _t(\phi (\theta _0,z_0+s)) \\ & = \left. \frac{d}{ds} \right|_{s=0} \left(\sqrt{1-(z_0+s+t)^2}\cos \theta _0, \sqrt{1-(z_0+s+t)^2} \sin \theta _0, z_0+s+t\right) \\ & = \left(-\frac{(z_0+t) \cos \theta _0}{\sqrt{1-(z_0+t)^2}},-\frac{(z_0+t) \sin \theta _0}{\sqrt{1-(z_0+t)^2}},1\right) \\ & = \left. \frac{\partial }{\partial z}\right|_{\Psi _t(p)}, \end{align*}$$

so flowing by \(\frac{\partial }{\partial z}\) also preserves the coordinate vector fields, but the coordinate vector fields do not have constant length on \(S^2\).

We see that

$$\begin{multline*} (\Psi _t^\ast g_{\Psi _t(p)})\left(a \left.\frac{\partial }{\partial \theta }\right|_p + b \left.\frac{\partial }{\partial z}\right|_p, c \left.\frac{\partial }{\partial \theta }\right|_p + d \left.\frac{\partial }{\partial z}\right|_p\right) = g_{\Psi _t(p)}\left(a \left.\frac{\partial }{\partial \theta }\right|_{\Psi _t(p)} + b \left.\frac{\partial }{\partial z}\right|_{\Psi _t(p)}, c \left.\frac{\partial }{\partial \theta }\right|_{\Psi _t(p)} + d \left.\frac{\partial }{\partial z}\right|_{\Psi _t(p)}\right) \\ = (1-(z_0+t)^2)ac + \frac{1}{1-(z_0+t)^2}bd, \end{multline*}$$

and hence

$$\begin{multline*} (\mathcal{L}_{\frac{\partial }{\partial z}} g)_p\left(\left.\frac{\partial }{\partial \theta }\right|_p,\left.\frac{\partial }{\partial \theta }\right|_p\right) = \left. \frac{d}{dt}\right|_{t=0} (\Psi _t^\ast g_{\Psi _t(p)})\left(\left.\frac{\partial }{\partial \theta }\right|_p,\left.\frac{\partial }{\partial \theta }\right|_p\right) = \left. \frac{d}{dt}\right|_{t=0} \left[(1-(z_0+t)^2)ac + \frac{1}{1-(z_0+t)^2}bd\right] \\ = -2z_0 ac + \frac{2z_0}{(1-z_0^2)^2}bd; \end{multline*}$$

that is,

\[ \mathcal{L}_{\frac{\partial }{\partial z}} g = -2z\, d\theta \otimes d\theta + \frac{2z}{(1-z^2)^2}dz \otimes dz, \]

which is no longer positive-definite (and hence not a Riemannian metric), but is still a symmetric \((0,2)\)-tensor field on \(S^2\) which captures how \(g\) changes as we flow in the \(\frac{\partial }{\partial z}\) direction.

One of the most important formulas in differential geometry is:

for any \(v_1, \dots , v_{k-1} \in T_pM\). This is sometimes also denoted by \(X \lrcorner \omega \). This is an antiderivation of degree \(-1\).

The reason Cartan’s magic formula is so important is that it is one of the most useful tools out there for computing exterior derivatives.

Notice, in particular, that we never had to work in local coordinates in Example 2.7.6, which is one of the great virtues of Cartan’s magic formula: it is one of the few tools that allows one to compute without using coordinates.

Hopefully this gives you some reason to believe that Cartan’s magic formula is useful, so let’s try to prove it:

so the formula holds for 0-forms.

Now consider a 1-form \(\omega \) which is the differential of a 0-form: \(\omega = du\). If \(Y \in \mathfrak {X}(M)\), then, by Definition 2.7.1,

using the definition of pullback (Definition 2.6.1). In turn, the chain rule tells us that the right hand side is equal to

by Lemma 1.3.4. Since \(Y\) is independent of \(t\), the (1-dimensional) chain rule tell us that this is equal to

So we have shown that \((\mathcal{L}_X\omega )_p(Y) = Y(X(u))(p)\).

On the other hand, \(\iota _X d\omega = \iota _X d(du) = 0\) and

by repeatedly using Lemma 1.3.4. Hence, the formula holds for differentials.

Finally, then, if \(\omega \in \Omega ^k(M)\) and we write \(\omega = \sum _I a_I dx_I\) in local coordinates, then

The \(k=0\) argument above implies that

On the other hand, again using the fact that the Lie derivative is a derivation, we have

since each \(dx_{i_j}\) is a differential.

Combining this with (2.6) and (2.7) implies that

as desired.

2.8 De Rham Cohomology

Recall that we have the cochain complex

where the subscript \(k\) indicates that \(d_k\) is the restriction of \(d\) to \(\Omega ^k(M)\).

De Rham cohomology is a homotopy invariant of manifolds, meaning that homotopy equivalent manifolds have identical de Rham cohomology groups.

If \(\omega \in \Omega ^k(M)\) and \(\eta \in \Omega ^\ell (M)\) are closed (i.e., \(d \omega = 0\) and \(d\eta = 0\)), meaning that they represent de Rham cohomology classes in \(H_{\text{dR}}^k(M)\) and \(H_{\text{dR}}^\ell (M)\), respectively, then

so \(d(\omega \wedge \eta )\) is also closed, and hence represents a class in \(H_{\text{dR}}^{k+\ell }(M)\). This means that

forms a (graded) ring, the de Rham cohomology ring, with the operation induced by \(\wedge \).

In fact, even more is true: the de Rham cohomology ring is isomorphic to the singular cohomology ring with real coefficients.

In Example 2.8.2, \(H^1(\mathbb {R}^2 -\{ 0\} ;\mathbb {R}) \cong H^1(S^1;\mathbb {R}) \cong \mathbb {R}\), so the 1-form \(\omega = \frac{x\, dy - y\, dx}{x^2 + y^2} \in \Omega ^1(\mathbb {R}^2 -\{ 0\} )\) is a generator of \(H_{\text{dR}}^1(\mathbb {R}^2-\{ 0\} )\).

We won’t give all the details of the proof of de Rham’s theorem, but the basic idea is that a closed form defines a linear functional on homology classes by integration. More precisely, the mapping \(I \colon \thinspace \! H_{\text{dR}}^k(M) \to H_k(M;\mathbb {R})^\ast \) is defined as follows: if \([\omega ] \in H_{\text{dR}}^k(M)\) is represented by the closed form \(\omega \in \Omega ^k(M)\), then \(I([\omega ]) \in H_k(M;\mathbb {R})^\ast \) is given by

for any \([C] \in H_k(M;\mathbb {R})\) which is represented by the cycle \(C \subset M\). It’s (hopefully!) clear that \(I([\omega ])\) is a linear functional (provided the integral here is anything sensible), but probably less obvious that \(I\) is well-defined and an isomorphism. Of course, to even make sense of the definition of \(I\), we need to talk about integration.

2.9 Integration on Manifolds

As suggested in Section 2.1, the point of defining differential forms as alternating, multilinear gadgets is because they transform in the right way.

2.10 Integration on Submanifolds

So far, we’ve only defined integrating an \(n\)-form over an \(n\)-dimensional manifold. However, notice that any open subset of an \(n\)-dimensional manifold is also an \(n\)-manifold, so we can actually integrate over arbitrary open subsets.

Indeed, given an \(n\)-manifold \(M\) and any \(\omega \in \Omega ^n(M)\), we get a (signed) measure \(m\) on open subsets of \(M\) defined by

When \(\omega \) is a volume form (meaning it is nowhere-vanishing), then \(m(A)\) will have the same sign for all open \(A\), and if the sign is positive then \(m\) will satisfy the axioms of a measure on the \(\sigma \)-algebra of Borel sets on \(M\); that is, it is a Borel measure. Moreover, if \(M\) is compact then \(\int _M \omega \) is finite, so \(m\) can be normalized to give a probability measure. Indeed, starting with a volume form is by far the most common way of defining probability measures on (orientable) manifolds.

However, this is not so useful for the integrals mentioned in Section 2.8 that show up in de Rham cohomology. After all, if \(N\) is a \(k\)-dimensional submanifold of the \(n\)-dimensional manifold \(M\) with \(k {\lt} n\) and \(\omega \in \Omega ^n(M)\), then \(\int _N \omega = 0\). ²⁵

Rather, for de Rham cohomology purposes, we need a way of defining \(\int _N \eta \) where \(\eta \in \Omega ^k(M)\). Intuitively, we want to “restrict” \(\eta \) to \(N\) here, but trying to write this intuition out directly leads to something quite complicated. But this is a place where all our effort in defining things in such a relatively abstract way pays off by giving us a simple and natural (though seemingly slightly indirect) way of defining our integral:

The point here is that \(i^\ast \eta \in \Omega ^k(N)\), and Definition 2.9.6 applies.

2.11 Stokes’ Theorem

At this point we have all the ingredients in place to prove (a version of) de Rham’s theorem: for a \(k\)-dimensional homology class \([C] \in H_k(M;\mathbb {R})\) represented by a submanifold \(C\), and a de Rham cohomology class \([\omega ]\) represented by a closed form \(\omega \in \Omega ^k(M)\), we at least know what

means.

However, to see that this is well-defined, we need to know that, if \([\omega _1] = [\omega _2]\), then \(\int _C \omega _1 = \int _C \omega _2\), and similarly that if \([C_1] = [C_2]\), then \(\int _{C_1}\omega = \int _{C_2}\omega \).

For the first, suppose \(\omega _1, \omega _2 \in \Omega ^k(M)\) represent the same de Rham cohomology class (i.e., the same element of \(H_{\text{dR}}^k(M) = \frac{\ker d}{\operatorname {im}d}\)), then \(\omega _1 - \omega _2 = d\eta \) for some \(\eta \in \Omega ^{k-1}(M)\). So we need to show that

Since \(C\) is a cycle, \(\partial C = \emptyset \), so this will follow from:

As for the second equality we need to show that the pairing \(I\) is well, defined, suppose \([C_1] = [C_2]\). This means ²⁶ that \(\partial E = C_2 - C_1\) for some \((k+1)\)-chain \(E\). But then for any closed \(\omega \in \Omega ^k(M)\),

by Theorem 2.11.1.

In other words, Stokes’ Theorem is the key to showing that the pairing \(I\) is well-defined. Of course, to really properly prove Theorem 2.8.3 we would also need to argue that every homology clas can be represented by a submanifold (or at least a finite union of submanifolds), which we’re not going to bother with.

To give a complete proof of Theorem 2.11.1, we would need to deveop the theory of manifolds with boundary. However, since we won’t really be using that theory for anything else, I don’t think it’s worth the effort. Instead, I’ll just give the part of the proof that doesn’t interact with the boundary. See Guillemin and Pollack  [ 17 ] for details on manifolds with boundary and a complete proof of Stokes’ Theorem.

We know that \(\phi ^\ast \omega \in \Omega ^{n-1}(U) \subset \Omega ^{n-1}(\mathbb {R}^n)\), so we can write

for some compactly supported functions \(f_i\), where \(d\check{x}_i\) means that \(dx_i\) is omitted and the signs are chosen so that

Since the \(f_i\) are compactly supported, we can extend their domain of definition to all of \(\mathbb {R}^n\) be defining them to be zero outside \(U\). Therefore,

By Fubini’s Theorem, we can interpret each term as an interated integral and do the integration in any order we like:

Now we’re thinking of the integrand \(\int _{-\infty }^\infty \frac{\partial f_i}{\partial x_i} \, dx_i\) as a function of \((x_1, \dots , \check{x}_i, \dots , x_n)\). Which function? Well, to any specific point \((a_1, \dots , \check{a}_i, \dots a_n)\) it assigns the number \(\int _{-\infty }^\infty g'(t)dt\) where \(g(t) = f_i(a_1, \dots , t, \dots , a_n)\). But since \(f_i\) is compactly supported, we know there exists \(T\) big enough that \(g(t) = 0\) for all \(t \geq T\).

Then the Fundamental Theorem of Calculus implies that

so we have

as desired.

To give the rest of the proof of Theorem 2.11.1, one needs to deal with the case \(\phi (U) \cap \partial M \neq \emptyset \). Again, see Guillemin and Pollack  [ 17 ] if you’re interested.

3.1 Some Motivation from de Rham Cohomology

Recall that we have the de Rham cohomology group \(H_{\text{dR}}^k(M)\), which tells us a lot about the topology of the manifold, but can in general be quite challenging to compute. After all, the \(\Omega ^k(M)\) are \(C^\infty (M)\)-modules, and hence infinite-dimensional as real vector spaces. If \(M\) is, e.g., compact then \(H_{\text{dR}}^k(M)\) will be a finite-dimensional real vector space, but it appears as the quotient of two infinite-dimensional vector spaces, so it can be somewhat challenging to compute.

However, when the manifold has lots of symmetry, computing the de Rham cohomology groups can actually become a finite-dimensional problem. In this case, a reasonable interpretation of “lots of symmetry” is “admits a transitive action of a Lie group,” resulting in a class of manifolds called homogeneous spaces. We haven’t defined Lie groups yet (though we will shortly), but the idea is that it’s a group which is also a manifold in a sensible way. At a more conceptual level, it’s a just a continuous family of symmetries of the manifold, where here “continuous” means they form a continuum.

The following statement says more precisely how de Rham cohomology becomes finite-dimensional in this setting, though don’t worry too much about terms in the statement that we haven’t defined yet.

Since, as we will see, \(G\)-invariant forms are determined by their behavior at a single point, this means that the space of \(G\)-invariant forms will be finite-dimensional (it can’t be bigger than \({\def\tmp {k} {\textstyle \bigwedge \! \! ^{k}} } (T_pM)^\ast \)), and hence computing de Rham cohomology becomes a finite-dimensional linear algebra problem.

Of course, the simplest possible example of a homogeneous space is a manifold that is itself a group, which acts by itself by (e.g. left) multiplication.

The point here is that de Rham cohomology simplies in the presence of symmetry.

Of course, this is a general principle: typically we expect lots of computations to simplify in the presence of symmetry. Lie groups provide the proper notion of (continuous) symmetries of manifolds, so if we want to understand symmetries and use them to simplify problems, we should probably care at least a little bit about Lie groups.

3.2 Lie Groups

The basic idea of a Lie group is that it is a group which is also a manifold, and that the group structure and the manifold structure are compatible. In this setting, compatibility is going to mean that, appropriately interpreted, the group operations are smooth maps.

Before actually stating the definition, the examples to have in mind are matrix groups: \(\operatorname {GL}_n(\mathbb {R})\), \(\operatorname {GL}_n(\mathbb {C})\), \(\operatorname {GL}_n(\mathbb {H})\), \(\operatorname {SL}_n(\mathbb {R})\), \(\operatorname {SL}_n(\mathbb {C})\), \(\operatorname {O}(n)\), \(\operatorname {SO}(n)\), \(\operatorname {U}(n)\), \(\operatorname {SU}(n)\), \(\operatorname {Sp}(n)\), the Heisenberg group \(\left\{ \begin{bmatrix} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1 \end{bmatrix} : a,b,c \in \mathbb {R}\right\} \), etc.

For example, \(\operatorname {GL}_n(\mathbb {C})\) is an open subset of \(\operatorname {Mat}_{n \times n}(\mathbb {C}) \cong \mathbb {C}^{n^2} \cong \mathbb {R}^{2n^2}\), so it is certainly a \(2n^2\)-dimensional manifold, and the entries of the product of two matrices can be written as (quadratic) polynomials in the entries of the terms, so this is smooth. Moreover, by the adjugate formula \(A^{-1} = \frac{1}{\det A} \operatorname {adj}(A)\), the entries of \(A^{-1}\) are rational functions of the entries of \(A\) with non-vanishing denominators, so taking the inverse of a matrix is also a smooth map.

Moreover, as we saw in Example 1.3.12, the unitary group \(\operatorname {U}(n)\) is the level set over a regular value of a smooth map defined on \(\operatorname {Mat}_{n \times n}(\mathbb {C})\), so this is also a manifold, and the same argument shows that the group operation and taking inverses are smooth.

The map that shows up in this definition is a slightly funny one; it’s maybe more common to require that the maps

• \(G \times G \to G\) given by \((g,h) \mapsto gh\) and

• \(G \to G\) given by \(g \mapsto g^{-1}\)

are smooth.

Notice that smoothness of \(L_g\) and \(R_g\) follows from smoothness of \((g,h) \mapsto gh^{-1}\).

3.3 Lie Algebras

One of the beautiful things about Lie groups is that the algebraic structure of the group and the geometric structure of the underlying manifold are closely related, so we can often learn things about the geometry by studying the algebra, or vice versa.

For example, the left multiplication map \(L_g \colon \thinspace \! G \to G\) is a diffeomorphism: we already know it’s smooth, and its inverse is clearly \(L_{g^{-1}}\), which is also smooth. Moreover, of \(e \in G\) is the identity element, then \(L_g(e) = g\), so the family of \(L_g\) diffeomorphisms will map the identity element of \(G\) to any other element of \(G\) that we like (equivalently, the action of \(G\) on itself by left multiplication is transitive). So even though from the algebraic perspective \(G\) has a distinguished point \(e\), from the geometric perspective all points look like all other points (since they all look like the identity).

Importantly, \(\left(dL_g\right)_e \colon \thinspace \! T_eG \to T_g G\) is a linear isomorphism, so we can explicitly identify every tangent space with the tangent space to the identity. Note, first of all, that this implies the tangent bundle is trivial: \(TG \cong G \times T_e G \cong G \times \mathbb {R}^n\), where \(n\) is the dimension of the group. Moreover, it tells us that it might be important to understand the tangent space to the identity.

Indeed, as we will shortly see, the tangent space to the identity has extra algebraic structure, and simply-connected Lie groups are completely determined by this algebraic structure.

The specific algebraic structure is that of a Lie algebra, which we’ve encountered before in Section 1.6 when talking about Lie brackets of vector fields, but now we give the formal definition:

By the last example, we know that for any Lie group \(G\) the collection of all vector fields \(\mathfrak {X}(G)\) is a Lie algebra. However, this is not the Lie algebra we’re after when we say the Lie algebra of \(G\): first, because Lie groups aren’t special in this regard (\(\mathfrak {X}(M)\) is a Lie algebra for any manifold), and second because \(\mathfrak {X}(M)\) is a (non-trivial) \(C^\infty (M)\)-module, and hence infinite-dimensional as a real vector space.

Building on ??, we can get a finite-dimensional sub-Lie algebra of \(\mathfrak {X}(G)\) by focusing on the left-invariant vector fields:

But then \(\operatorname {ev}_e(X) = v\), so \(\operatorname {ev}_e\) is surjective.

On the other hand, \(\operatorname {ev}_e\) is also injective: if \(\operatorname {ev}_e(X) = \operatorname {ev}_e(Y)\) for \(X,Y \in \mathfrak {g}\), then for all \(g \in G\) we have

by left-invariance of \(X\) and \(Y\), so we conclude that \(\operatorname {ev}_e(X) = \operatorname {ev}_e(Y)\) implies \(X=Y\).

2 is straightforward but annoying, so we skip it.

3 is a consequence of 3.3.8 stated below, and then 4 is immediate given that the Lie bracket of vector fields is alternating and satisfies the Jacobi identity (which you proved in HW 1 #4).

In the proof of Theorem 3.3.7 3, we apply this lemma with \(M=N=G\) and \(f = L_g\).

Theorem 3.3.7 tells us that any Lie group \(G\) has an associated finite-dimensional Lie algebra \(\mathfrak {g}\) of left-invariant vector fields; this is usually called the Lie algebra of \(G\), typically denoted with a lowercase fraktur version of the name of the group, but sometimes also \(\operatorname {Lie}(G)\).

Notice, in particular, that we can also identify \(\mathfrak {g}\) with \(T_eG\) using the evaluation map, so we could also interpret the Lie bracket as being an operation on this tangent space.

3.4 Matrix Groups

You might worry that this excludes groups consisting of complex or quaternionic matrices, but all such groups can be represented as subgroups of some real general linear group. Specifically, define the map \(\rho _1 \colon \thinspace \! \mathbb {C}\to \operatorname {Mat}_{2 \times 2} (\mathbb {R})\) by

This is the right definition to turn multiplication of complex numbers into linear transformations of \(\mathbb {R}\). After all, if \(z= x+iy\), then for \(w = u + i v\) we have

and

In turn, we can get a map \(\rho _n \colon \thinspace \! \operatorname {Mat}_{n \times n}(\mathbb {C}) \to \operatorname {Mat}_{2n \times 2n}(\mathbb {R})\) by just applying \(\rho _1\) to each entry, resulting in a real matrix built out of \(2 \times 2\) blocks of the form \(\begin{bmatrix} x & -y \\ y & x \end{bmatrix}\). For example

It’s not hard to show that \(\rho _n\) preserves invertibility, so \(\rho _n(\operatorname {GL}_n(\mathbb {C})) \subset \operatorname {GL}_{2n}(\mathbb {R})\), showing that \(\operatorname {GL}_n(\mathbb {C})\) and all of its closed subgroups are matrix groups.

Similarly, we can define \(\psi _1 \colon \thinspace \! \mathbb {H}\to \operatorname {Mat}_{2 \times 2}(\mathbb {C})\) by

or, written out more completely,

And then we can extend to a map \(\psi _n \colon \thinspace \! \operatorname {Mat}_{n \times n}(\mathbb {H}) \to \operatorname {Mat}_{2n \times 2n}(\mathbb {C})\) as you’d expect. For example,

Again, \(\psi _n(\operatorname {GL}_n(\mathbb {H})) \subset \operatorname {GL}_{2n}(\mathbb {C})\), and of course by composing \(\psi _n\) with \(\rho _{2n}\) we can represent any quaternionic matrix group as a subgroup of \(\operatorname {GL}_{4n}(\mathbb {R})\).

The nice thing about matrix groups is that the bracket operations on their Lie algebras are particularly simple.

We’re going to prove this by writing out the matrix multiplication in terms of coordinates; i.e., if \(A,B \in \operatorname {Mat}_{n \times n}(\mathbb {R})\), then

But now, if we’re thinking of our matrices as tangent vectors at the identity, then they’re really differential operators, so we need a way of interpreting these coordinates in that language.

To that end, by analogy with what we did in Section 2.4.1, define \(x_{ij} \colon \thinspace \! \operatorname {Mat}_{n \times n}(\mathbb {R}) \to \mathbb {R}\) by \(x_{ij}(A) = A_{ij}\), the \((i,j)\) entry of \(A\). Then if \(V \in T_I \operatorname {GL}_n(\mathbb {R}) \cong \operatorname {Mat}_{n \times n}(\mathbb {R})\), we can write \(V = \alpha '(0)\) where \(\alpha (t) = I + tV\), which stays in \(\operatorname {GL}_n(\mathbb {R})\) for small enough \(t\). Therefore, using Definition 1.2.1

So this gives a way of translating back and forth between elements of \(T_I\operatorname {GL}_n(\mathbb {R})\) as differential operators and as arrays of numbers.

Next, recall the evaluation map \(\operatorname {ev}_I \colon \thinspace \! \mathfrak {gl}_n(\mathbb {R}) \to T_I \operatorname {GL}_n(\mathbb {R})\) given by \(\operatorname {ev}_I(X) = X(I)\), the tangent vector at \(I\) determined by the left-invariant vector field \(X\). In what follows, to avoid a surplus of parentheses, we will sometimes write \(X(I)\) as \(X_I\).

We saw in the proof of Theorem 3.3.7 that \(\operatorname {ev}_I\) is a vector space isomorphism, so all that remains to do is to show that it is a Lie algebra homomorphism; that is, for any left-invariant vector fields \(X,Y\) on \(\operatorname {GL}_n(\mathbb {R})\) (i.e., elements of \(\mathfrak {gl}_n(\mathbb {R})\)), we need to show that

where on the right hand side we’re just multiplying matrices.

Notice, first of all, that for any \(g,h \in \operatorname {GL}_n(\mathbb {R})\),

in other words,

We’re going to look now at the function \(Y(x_{ij})\), which we can specify by saying how it evaluates at each \(g \in \operatorname {GL}_n(\mathbb {R})\). By left-invariance of \(Y\) together with (None) and (None),

This (and the corresponding statement for \(X\)) gives us the necessary tool to determine the \((i,j)\) component of \(\operatorname {ev}_I([X,Y])\):

by linearity of the operators \(X_I\) and \(Y_I\). Of course, \(Y(I)_{kj}\) and \(X(I)_{kj}\) are constant, so the Leibniz rule gives us

where, as a reminder, I’m freely switching between the notations \(X(I)\) and \(X_I\) for the tangent vector at \(I\) determined by the vector field \(X\). So this shows that the left and right sides of the equation in the statement of the theorem agree componentwise.

This now justifies the computational simplifications we did in looking at the Lie algebras of \(\operatorname {U}(d)\) (Example 1.6.8), \(\operatorname {SO}(d)\) (Example 1.6.10), and \(\operatorname {Aff}(\mathbb {R})\) (Example 3.3.9).

For example, \(T_I\operatorname {SO}(3)\) consists of the \(3 \times 3\) skew-symmetric matrices, which we saw in Example 1.6.10 is isomorphic to \((\mathbb {R}^3,\times )\).

We’ve seen in Example 1.5.3 that \(T_I\operatorname {U}(d)\) consists of the skew-Hermitian \(d \times d\) matrices; I claim without proof that the determinant-1 condition translates to a trace-0 condition, so \(T_I\operatorname {SU}(d)\) is the space of traceless skew-Hermitian matrices.

Finally, thinking of \(\operatorname {Sp}(1) \cong S^3\) as the \(1 \times 1\) quaternionic matrices which preserve the symplectic inner product (i.e., the quaternionic analog of \(\operatorname {O}(d)\) or \(\operatorname {U}(d)\)), the same sort of argument that gave the tangent spaces at the identity of \(\operatorname {O}(d)\) and \(\operatorname {U}(d)\) will imply that \(T_I \operatorname {Sp}(1)\) consists of quaternionic skew-Hermitian \(1 \times 1\) matrices; in other words, the single entry must be a purely imaginary quaternion \(ai + bj + ck\).

We can fairly easily write down bases for \(\mathfrak {so}(3)\), \(\mathfrak {su}(2)\), and \(\mathfrak {sp}(1)\):

If we use \(\{ A_{1,\mathfrak {g}},A_{2,\mathfrak {g}},A_{3,\mathfrak {g}}\} \) to denote the basis for \(\mathfrak {g} \in \{ \mathfrak {so}(3), \mathfrak {su}(2),\mathfrak {sp}(1)\} \), then it is easy to check that

For example,

and

Therefore, for \(\{ \mathfrak {g}, \mathfrak {g}'\} \subset \{ \mathfrak {so}(3), \mathfrak {su}(2),\mathfrak {sp}(1)\} \), linearly extending the identifications \(A_{i,\mathfrak {g}} \mapsto A_{i,\mathfrak {g}'}\) defines a Lie algebra isomorphism.

Proposition 3.4.3 is an example of what is sometimes called an accidental isomorphism, which in this case boil down to the fact that there just aren’t that many 3-dimensional Lie algebras. In general \(\mathfrak {su}(n+1)\), \(\mathfrak {so}(2n+1)\), and \(\mathfrak {sp}(n)\) belong to three different infinite families of Lie algebras: they are \(A_n\), \(B_n\), and \(C_n\), respectively. In this language, Proposition 3.4.3 says that \(A_1 \cong B_1 \cong C_1\).

3.5 Adjoint and Coadjoint Representations

For each \(g \in G\), let \(C_g \colon \thinspace \! G \to G\) denote conjugation by \(g\), namely \(C_g(h) = ghg^{-1}\). Notice, in particular, that \(C_g(e) = geg^{-1} = e\) for all \(g \in G\), so these maps are all diffeomophisms fixing the identity, and hence \(\left(dC_g\right)_e \colon \thinspace \! T_e G \to T_e G\) for all \(g \in G\).

When \(G\) is a matrix group it’s easy to get a formula for \(\left(dC_g\right)_e\): since \(C_g = L_g \circ R_{g^{-1}}\), we know that \(dC_g = d(L_g \circ R_{g^{-1}}) = dL_g \circ dR_{g^{-1}}\), so it suffices to find nice formulas for \(dL_g\) and \(dR_{g^{-1}}\). For example, if \(v \in T_h G\) and \(v = \alpha '(0)\) for some smooth curve \(\alpha (t)\) with \(\alpha (0)=h\), then

where the operation here is matrix multiplication, which makes sense when we intepret \(T_hG \subset T_h \operatorname {GL}_n(\mathbb {R}) = \operatorname {Mat}_{n \times n}(\mathbb {R})\).

The analogous statement is true for \(dR_{g^{-1}}\), so we see that, on matrix groups, \(\left(dC_g\right)_ev = gvg^{-1}\); that is, this is just the conjugation action of the group on its tangent space at the identity.

Of course, we can identify \(T_eG\) with the Lie algebra \(\mathfrak {g}\), so we can also think of \(\left(dC_g\right)_e\) as a map \(\mathfrak {g} \to \mathfrak {g}\), but in this guise we usually call the map \(\operatorname {Ad}_g\) rather than \(\left(dC_g\right)_e\); that is, we have \(\operatorname {Ad}_g \colon \thinspace \! \mathfrak {g} \to \mathfrak {g}\), called the adjoint action of \(g \in G\) on \(\mathfrak {g}\). Said yet another way, the following is well-defined:

Notice that \(\operatorname {Ad}\) is a Lie group homomorphism: since

we see that \(C_{gh} = C_g \circ C_h\), so

One of the key virtues of the adjoint representation is that it defines a homomorphism from any Lie group to a matrix group. Moreover, \(\ker \operatorname {Ad}= Z(G)\), the center of \(G\), so the first isomorphism theorem implies that \(\operatorname {Ad}(G) \cong G/Z(G)\); in particular, this shows that any centerless Lie group can be realized as a matrix group.

In various circumstances, some of which we may encounter later, it is more convenient to dualize, meaning we want to look at the induced action on \(\mathfrak {g}^\ast \).

Given \(g \in G\), the map \(\operatorname {Ad}_g \colon \thinspace \! \mathfrak {g} \to \mathfrak {g}\) has, in the spirit of Section 2.9.1, an associated dual map \(\left(\operatorname {Ad}_g\right)^\ast \colon \thinspace \! \mathfrak {g}^\ast \to \mathfrak {g}^\ast \) defined by

for any \(\xi \in \mathfrak {g}^\ast \) and any \(X \in \mathfrak {g}\), but this does not define a homomorphism \(G \to \operatorname {Aut}(\mathfrak {g}^\ast )\): for any \(g,h \in G\) and \(\xi \in \mathfrak {g}^\ast \),

for any \(X \in \mathfrak {g}\). In other words, \(\left(\operatorname {Ad}_{gh}\right)^\ast = \left(\operatorname {Ad}_h\right)^\ast \circ \left(\operatorname {Ad}_g\right)^\ast \), which is not generally equal to \(\left(\operatorname {Ad}_g\right)^\ast \circ \left(\operatorname {Ad}_h\right)^\ast \).

To fix this and get an honest Lie group homomorphism, we need to twist by the involution \(g \mapsto g^{-1}\) (in fancy categorical language, I believe this is the canonical isomorphism of \(G\) with \(G^{\text{op}}\) and the dual map really defines a homomorphism \(G^{\text{op}} \to \operatorname {Aut}(\mathfrak {g}^\ast )\)):

One virtue of the coadjoint action is that each orbit (called a coadjoint orbit) has a natural symplectic structure.

3.6 Lie Algebra Homomorphisms

We’ve just seen that a Lie group always acts on its Lie algebra by the adjoint action. Now we start to develop another connection between Lie groups and Lie algebras, one already hinted at:

One way to interpret this is that Lie group homomorphisms are (essentially) uniquely determined by their behavior in arbitrarily small neighborhoods of the identity (since they’re determined by their linearizations at the identity). A more or less immediate corollary of the theorem is that simply-connected Lie groups are uniquely determined by their Lie algebras:

A first step in the proof of Theorem 3.6.1 is to show that Lie group homomorphisms induce Lie algebra homomorphisms:

Before proving this, it’s probably a good idea to pause for a moment and decipher notation. By definition, \(\mathfrak {g}\) is the collection of left-invariant vector fields on \(G\), so if \(X\) is a left-invariant vector field on \(G\), then \(d\phi (X)\) is the vector field on \(H\) given by \(\left(d\phi \right)_gX(g)\) at each point \(\phi (g)\) in the image of \(\phi \), and then extended to a left-invariant vector field on all of \(H\).

where we used the Chain Rule in the first and third equalities and the left-invariance of \(X\) in the last equality. The second equality follows because \(\phi \) is a Lie group homomorphism in the second; more explicitly, for any \(g_1,g_2 \in G\),

so we see that \(L_{\phi (g_1)} \circ \phi = \phi \circ L_{g_1}\) and the second equality in (3.5) follows.

Thus, we have shown that \(d\phi \) really maps \(\mathfrak {g}\) to \(\mathfrak {h}\). To see that it is an Lie algebra homomorphism, recall from Lemma 3.3.8 that, for \(X,Y \in \mathfrak {g}\) and \(g \in G\),

or, more poetically, \(d\phi [X,Y] = [d\phi X, d\phi Y]\), which is what it means for \(d\phi \) to be a Lie algebra homomorphism.

Proposition 3.6.3 shows that Lie group homomorphisms induce Lie algebra homomorphisms, and so the expression \(d\phi = \psi \) in Theorem 3.6.1 makes sense. In order to prove the rest of Theorem 3.6.1, we’ll need to make an extended detour into the theory of distributions (which are also useful for other things).

3.7 Distributions and the Frobenius Theorem

Think back to Section 1.2 when we defined tangent spaces and the tangent bundle. The upshot of that construction (Definition 1.2.1) was that we assigned a vector space to each point on the manifold, and that the elements of each tangent space were all possible velocities of smooth curves passing through the point.

Now, in a variety of situations, we may be particularly intgerested in only certain types of curves passing through each point.

Example 3.7.1

Consider a homogeneous, first-order differential equation of the form \(F(x,z,z') = 0\), where \(z=z(x)\) is a function of \(x\). Solutions correspond to curves \(x \mapsto (x,z(x),z'(x))\), which we can think of as living in \(\mathbb {R}^3 = \mathbb {R}^2 \times \mathbb {R}\) with coordinates \((x,z,p)\). But such curves cannot have arbitrary velocities (or tangent vectors): specifically, \(dz = z'\, dx\), so the tangents to the curve must always lie in the kernel of the 1-form

\[ \alpha = dz - p\, dx. \]

This 1-form is an example of a contact form, first introduced in the late 19th century by Sophus Lie  [ 23 ] . He called this a contact element or a line element because \(dz - p\, dx = 0\) can be interpreted as the differential form of the equation of a line of slope \(p\) passing through the point \((x,z) \in \mathbb {R}^2\).

Exercise 3.7.2

Show that \(\ker \alpha = \operatorname {span}\left\{ \frac{\partial }{\partial x} + p \frac{\partial }{\partial z}, \frac{\partial }{\partial p}\right\} \).

Therefore, \(\ker \alpha \) determines a 2-dimensional subspace of each tangent space, visualized in Figure 3.1.

Solutions to differential equations of the form \(F(x,z,z')=0\) produce curves in \((x,z,p)\)-space which are everywhere tangent to the plane determined by \(\ker \alpha \); these are often called Legendrian curves. This formalism is the basis of an area of mathematics called contact geometry or contact topology, which has recently been very useful in the study of 3-manifolds, but which historically was developed because of its relevance to thermodynamics and geometric optics; see Geiges’s historical survey  [ 14 ] for an introduction to this area.

The plane field described above and illustrated in Figure 3.1 is an example of a distribution:

The algebraic condition of involutivity is closely related to the geometric condition of having an integral manifold:

using Lemma 3.3.8. So \(\mathcal{D}\) is involutive.

The much more amazing fact is that the converse is true, so we can determine existence integral manifolds by checking involutivity. This is the content of:

The idea behind the theorem is the following: suppose \(\mathcal{D}\) is (locally) spanned by the vector fields \(V_1, \dots , V_k\). Remember that the \(V_i\) are really first-order differential operators on \(M\), so the set \(\{ V_1, \dots , V_k\} \) determines a first-order homogeneous system of PDEs

If \(u \colon \thinspace \! M \to \mathbb {R}\) is a solution to this system of PDEs, then the level sets of \(u\) are tangent to \(\operatorname {span}\{ V_1, \dots , V_k\} \). Frobenius’ original question was: when are there solutions \(u_1, \dots , u_{n-k}\) so that the gradients \(\nabla u_1, \dots , \nabla u_{n-k}\) are linearly independent? (There certainly can’t be more than \(n-k\) solutions, since the gradients will be normal to the level sets and the orthogonal complement of \(\operatorname {span}\{ V_1, \dots , V_k\} \) is \((n-k)\)-dimensional.)

When there exist such \(u_1, \dots , u_{n-k}\), then the level sets of the product map \((u_1, \dots , u_{n-k}) \colon \thinspace \! M \to \mathbb {R}^{n-k}\) will necessarily be integral manifolds of \(\mathcal{D}\).

1. \(\vec{0} \in U \subset \mathbb {R}^n\)

2. \(\phi (\vec{0})=p\)

3. For \(\vec{a} = (a_1, \dots , a_n) \in U\), the set \(\{ q \in M : \phi ^{-1}(q)_{k+1} = a_{k+1}, \dots , \phi ^{-1}(q)_n = a_n\} \) is an integral manifold of \(\mathcal{D}\).

Since this is a local argument, we may as well assume \(M \subset \mathbb {R}^n\) is an open set, \(p = \vec{0}\), and, after rotation, \(\mathcal{D}(p) = \mathcal{D}(\vec{0}) = \operatorname {span}\left\{ \frac{\partial }{\partial x_1} , \dots , \frac{\partial }{\partial x_k}\right\} \) is the \(k\)-plane spanned by the first \(k\) coordinate directions.

Let \(\pi \colon \thinspace \! \mathbb {R}^n \to \mathbb {R}^k\) be projection onto the first \(k\) coordinates, so that \(\left. (d\pi )_{\vec{0}}\right|_{\mathcal{D}} \colon \thinspace \! \mathcal{D}(\vec{0}) \to T_{\vec{0}} \mathbb {R}^k \cong \mathbb {R}^k\) is an isomorphism, and hence \(\left. d\pi \right|_{\mathcal{D}}\) is injective on some neighborhood \(U\) of \(\vec{0}\).

Therefore, in \(U\) there exist unique \(V_1, \dots , V_k \in \mathcal{D}\) so that \((d\pi )_q V_i = \left. \frac{\partial }{\partial x_i} \right|_{\pi (q)}\) for all \(q \in U\) and for all \(i \in \{ 1 , \dots , k\} \).

Since \(\mathcal{D}\) is involutive, \([V_i, V_j ] \in \mathcal{D}\) for all \(1 \leq i, j \leq k\). On the other hand,

so \([V_i,V_j]= 0\) since \(d\pi \) is an isomorphism.

But then we can find local coordinates \(y_1, \dots , y_n\) so that \(V_i = \frac{\partial }{\partial y_i}\) on \(U\) and the sets

are integral manifolds of \(\mathcal{D}\).

Example 3.7.11

Let \(M = \mathbb {R}^2 - \{ \vec{0}\} \), let \(X(u,v) = -v \frac{\partial }{\partial x} + u \frac{\partial }{\partial y}\), and let \(\mathcal{D} = \operatorname {span}\{ X\} \). \(\mathcal{D}\) is certainly involutive since \([X,X]=0\) by anticommutativity of the Lie bracket, so there must be (local) integral manifolds through each point in \(M\). Of course, this example is easy: the integral manifolds are just concentric circles, as we see in Figure 3.2.

In fact, anticommutativity of the Lie bracket implies that all one-dimensional distributions are involutive, and hence integrable. Said another way: given any line field on a manifold and any point in the manifold, there is a curve passing through the point whose velocity vectors are everywhere tangent to the line field.

The Frobenius theorem we’ve proved is a local result (after all, there’s no guarantee that the integral manifold we get has diameter \({\gt} \epsilon \)), but it turns out that the local pieces glue together nicely:

Example 3.7.14

The foliation of \(\mathbb {R}^2 - \{ \vec{0}\} \) corresponding to the distribution \(\mathcal{D} = \operatorname {span}\left\{ -y \frac{\partial }{\partial x} + x \frac{\partial }{\partial y}\right\} \) from Example 3.7.11 is precisely the disjoint union of concentric circles shown in Figure 3.3.

Example 3.7.15

The foliation of \(\mathbb {R}^3\) corresponding to the involutive distribution \(\mathcal{D} = \operatorname {span}\left\{ \frac{\partial }{\partial x}, \frac{\partial }{\partial y} \right\} \) is the collection of planes parallel to the \(xy\)-plane, as shown in Figure 3.4.

3.8 Distributions and Differential Ideals

3.8.2 Differential Ideals

Now we formalize our approach to describing distributions as kernels of differential forms.

Definition 3.8.5

Suppose \(\mathcal{D}\) is a \(d\)-dimensional distribution on an \(n\)-dimensional manifold \(M\) and define \(\mathcal{I}(\mathcal{D}) \subset \Omega ^\ast (M)\) to be the ring of differential forms which annihilate \(\mathcal{D}\):

\[ \mathcal{I}(\mathcal{D}) := \bigcup _k \left\{ \omega \in \Omega ^k(M) : \omega (X_1, \dots , X_k) = 0 \text{ for all } X_1, \dots , X_k \in \mathcal{D} \right\} . \]

If \(\omega _1, \omega _2 \in \mathcal{I}(\mathcal{D})\), then \(\omega _1 + \omega _2 \in \mathcal{I}(\mathcal{D})\) and \(\omega _1 \wedge \omega _2 \in \mathcal{I}(\mathcal{D})\), so this really is a ring.

In fact, if \(\omega \in \mathcal{I}(\mathcal{D}) \cap \Omega ^k(M)\) and \(\eta \in \Omega ^\ell (M)\), then

\[ \omega \wedge \eta (X_1, \dots , X_{k+\ell }) = \sum _{\sigma \in S(k,\ell )} \operatorname {sgn}(\sigma ) \omega (X_{\sigma (1)}, \dots , X_{\sigma (k)}) \eta (X_{\sigma (k+1)}, \dots , X_{\sigma (k+\ell )}) = 0, \]

so \(\omega \wedge \eta \in \mathcal{I}(\mathcal{D})\), which means that \(\mathcal{I}(\mathcal{D})\) is actually an ideal, usually called either the annihilator ideal of \(\mathcal{D}\) or just the annihilator of \(\mathcal{D}\).

We’ve seen before in the proof of Theorem 3.7.10 that any point \(p \in M\) is contained in a coordinate chart \((U,\phi )\) with local coordinates \(x_1, \dots , x_n\) so that \(\mathcal{D}(p) = \operatorname {span}\left\{ \frac{\partial }{\partial x_1}, \dots , \frac{\partial }{\partial x_d} \right\} \). If \(\xi _1, \dots , \xi _n\) are the dual 1-forms (i.e., \(\xi _i = dx_i\)), then \(\xi _{d+1}, \dots , \xi _n\) are independent and generate \(\mathcal{I}(\mathcal{D})\), which proves:

Lemma 3.8.6

For any \(d\)-dimensional distribution \(\mathcal{D}\), the annihilator ideal \(\mathcal{I}(\mathcal{D})\) is locally generated by \(n-d\) independent \(1\)-forms \(\xi _{d+1},\dots , \xi _n\).

On the other hand:

Lemma 3.8.7

If \(\mathcal{I} \subset \Omega ^\ast (M)\) is an ideal locally generated by \(n-d\) independent \(1\)-forms, then there is a \(d\)-dimensional distribution \(\mathcal{D}\) on \(M\) so that \(\mathcal{I} = \mathcal{I}(\mathcal{D})\).

Proof.

Let \(\omega _{d+1}, \dots , \omega _n\) be the independent 1-forms generating \(\mathcal{I}\) in a neighborhood of \(p \in M\). Then define \(\mathcal{D}(p) \subset T_pM\) to be the \(d\)-dimensional subspace annihilated by the linear functionals \(\left(\omega _{d+1}\right)_p, \dots , \left(\omega _n\right)_p\). Then \(\mathcal{D} = \bigcup _{p \in M} \mathcal{D}(p)\) is the desired distribution.

Now we can give an alternative version of the Frobenius theorem in terms of annihilator ideals:

Theorem 3.8.8 (Frobenius Theorem, version 2)

A distribution \(\mathcal{D}\) on \(M\) is involutive (and hence integrable) if and only if \(\mathcal{I}(\mathcal{D})\) is closed under taking exterior derivatives: \(d(\mathcal{I}(\mathcal{D})) \subseteq \mathcal{I}(\mathcal{D})\).

Definition 3.8.9

An ideal \(\mathcal{I} \subset \Omega ^\ast (M)\) is called a differential ideal if \(d(\mathcal{I}) \subseteq \mathcal{I}\).

So a restatement of Theorem 3.8.8 is that a distribution is involutive if and only if its annihilator ideal is a differential ideal.

Proof.

[Proof of Theorem 3.8.8] Locally choose \(\xi _1, \dots , \xi _n\) that span \(\left(T_pM\right)^\ast = {\def\tmp {1} {\textstyle \bigwedge \! \! ^{1}} }\left(\left(T_pM\right)^\ast \right)\) so that \(\xi _{d+1}, \dots , \xi _n\) generate \(\mathcal{I}(\mathcal{D})\) in a neighborhood of \(p \in M\). Let \(X_1, \dots , X_n\) be the dual vector fields so that \(\xi _i(X_j) = \delta _{ij}\). Then \(X_1, \dots , X_d\) span \(\mathcal{D}\) and \(\mathcal{D}\) is involutive if and only if \([X_i, X_j] \in \mathcal{D}\) for all \(1 \leq i, j \leq d\).

But now we get to use Cartan’s Magic Formula (Theorem 2.7.5), which says that \(\mathcal{L}_X \omega = \iota _X d\omega + d \iota _X \omega \). Then for any \(1\)-form \(\omega \) and any \(X,Y \in \mathfrak {X}(M)\), this implies that

$$\begin{align*} d \omega (X,Y) = (\iota _X d\omega )(Y) & = (\mathcal{L}_X \omega )(Y)-(d \iota _X \omega )(Y) \\ & = \mathcal{L}_X(\omega (Y))-\omega (\mathcal{L}_X(Y))-(d\omega (X))(Y) \\ & = X(\omega (Y))-\omega ([X,Y]) - Y(\omega (X)), \end{align*}$$

since \(\omega (X)\) and \(\omega (Y)\) are functions.

We record this as a lemma:

Lemma 3.8.10

Suppose \(\omega \in \Omega ^1(M)\) and \(X,Y \in \mathfrak {X}(M)\). Then

\[ d \omega (X,Y) = X(\omega (Y))- Y(\omega (X))-\omega ([X,Y]). \]

This implies that, if \(1 \leq i, j \leq d\) and \(d+1 \leq k \leq n\),

\[ d\xi _k(X_i,X_j) = X_i(\xi _x(X_j)) - X_j(\xi _k(X_i)) - \xi _k([X_i,X_j]) = X_i(0) - X_j(0) - \xi _k([X_i,X_j]) = - \xi _k([X_i,X_j]). \]

This is zero for all such \(i,j,k\) if and only if \(\mathcal{D}\) is involutive. So \(\mathcal{I}(\mathcal{D})\) is a differential ideal if and only if \(\mathcal{D}\) is involutive.

Example 3.8.11

Recall yet again Section 1.8 and our left-invariant vector fields \(X,Y,Z\) and dual left-invariant 1-forms \(\alpha , \beta , \gamma \) on \(S^3\). Let \(\zeta = \operatorname {span}\{ Y,Z\} = \ker \alpha \). Recall from Example 3.7.6 that \([Y,Z] = 2X \notin \zeta \), so \(\zeta \) is not involutive, which by the original Frobenius Theorem (Theorem 3.7.10) implies that \(\zeta \) is not integrable. We can also see this using Theorem 3.8.8: certainly \(\alpha \in \mathcal{I}(\zeta )\), but as we saw in Example 2.7.6,

\[ d\alpha = -2 \beta \wedge \gamma \]

which is not in \(\mathcal{I}(\zeta )\) since \(-2\beta \wedge \gamma (Y,Z) = -2 \neq 0\), so \(\mathcal{I}(\zeta )\) is not a differential ideal.

We (finally) have all the tools in place to start using this distribution machinery to prove Theorem 3.6.1, which said that Lie algebra homomorphisms induce Lie group homomorphisms.

Here’s an intermediate result:

Theorem 3.8.12

Let \(G\) be a connected Lie group and let \(f_1, f_2 \colon \thinspace \! G \to H\) be Lie group homomorphisms so that the Lie algebra homomorphisms \(df_1, df_2 \colon \thinspace \! \mathfrak {g} \to \mathfrak {h}\) are identical. Then \(f_1 \equiv f_2\).

Let’s start trying to prove this and see where we run into trouble. Since \(df_1 = df_2\) as maps \(\mathfrak {g} \to \mathfrak {h}\), the dual maps \(f_1^\ast = f_2^\ast \) as maps \(\mathfrak {h}^\ast \to \mathfrak {g}^\ast \), where we identify \(\mathfrak {g}^\ast \) and \(\mathfrak {h}^\ast \) with the left-invariant 1-forms on \(G\) and \(H\), respectively. ³⁰

Suppose \(\omega _1 , \dots , \omega _n\) is a basis for the left-invariant forms on \(H\). Then we know that \(f_1^\ast \omega _i = f_2^\ast \omega _i\) for all \(i = 1, \dots , n\), and the question is whether this is enough to show that \(f_1 \equiv f_2\).

Think of this as a special case of a more general problem:

Problem 3.8.13

Suppose we have manifolds \(M\) and \(N\) of dimensions \(m\) and \(n\), respectively, and we have a basis \(\omega _1, \dots , \omega _n\) for the 1-forms on \(N\), as well as a preferred collection \(\alpha _1, \dots , \alpha _n\) of 1-forms on \(M\). Is there a (preferably unique) map \(f \colon \thinspace \! M \to N\) so that

\[ f^\ast \omega _i = \alpha _i \]

for all \(i = 1, \dots , n\)?

Example 3.8.14

Let \(M = \mathbb {R}\), let \(N = \mathbb {R}_{{\gt}0}\) be the positive reals, let \(\omega = \frac{1}{y} \, dy \in \Omega ^1(N)\), and let \(\alpha = dx \in \Omega ^1(M)\). If there is some \(f \colon \thinspace \! M \to N\) so that \(f^\ast \omega = \alpha \), then it must be the case that, for each \(r \in \mathbb {R}\),

\[ 1 = dx_r\left(\frac{\partial }{\partial x} \right) = \alpha _r\left(\frac{\partial }{\partial x} \right) = (f^\ast \omega )_r\left(\frac{\partial }{\partial x} \right) = \omega _{f(r)} \left(df_r\frac{\partial }{\partial x} \right) = \omega _{f(r)} \left((f\circ \gamma )'(0)\right), \]

where \(\gamma (0) = r\) and \(\gamma '(0) = \frac{\partial }{\partial x}\); for example, \(\gamma (t) = r+t\). In this case, \((f\circ \gamma )(t) = f(r+t)\), so \((f\circ \gamma )'(0) = f'(r) \frac{\partial }{\partial y}\), and hence we see that

\[ 1 = \omega _{f(r)} \left((f\circ \gamma )'(0)\right) = \omega _{f(r)}\left( f'(r) \frac{\partial }{\partial y}\right) = \frac{1}{f(r)} dy \left( f'(r) \frac{\partial }{\partial y}\right) = \frac{1}{f(r)}f'(r). \]

This can be solved, for example with \(f_c(r) = c e^r\) for any positive constant \(c {\gt} 0\). If we specify that \(f(0) = 1\), then that fixes \(c=1\) and we have the unique solution \(f(r) = e^r\).

To make this more geometric, for each \(c {\gt} 0\) consider the graph of \(f_c\), shown in Figure 3.5:

\[ \Gamma _c := \left\{ \left(r, f_c(r) \right) : r \in \mathbb {R}\right\} = \left\{ \left(r, ce^r \right) : r \in \mathbb {R}\right\} . \]

Notice that the tangent line to a point \((r, ce^r) \in \Gamma _c\) is spanned by the vector \((1,ce^r) = \frac{\partial }{\partial x} + c e^r \frac{\partial }{\partial y}\). If we write \((r, ce^r)\) as \((x,y)\), then this vector is just \(\frac{\partial }{\partial x} + y \frac{\partial }{\partial y}\), which defines a non-vanishing vector field (and hence a line field or 1-dimensional distribution) on all of the upper half-plane \(\mathbb {R}\times \mathbb {R}_{{\gt}0}\). And for each \(c\) the graph \(\Gamma _c\) is an integral manifold of this distribution. See Figure 3.6.

Finally, we see that the annihilator ideal of this distribution is generated by the 1-form dual to \(\frac{\partial }{\partial x} + y \frac{\partial }{\partial y}\), namely \(\mu = dx - \frac{1}{y} \, dy\). But now this is a form we could have constructed on \(M \times N = \mathbb {R}\times \mathbb {R}_{{\gt}0}\) without knowing anything at all about \(f\): if \(\pi _1\) and \(\pi _2\) are the obvious projections onto the first and second factors, then \(\mu = \pi _1^\ast \alpha - \pi _2^\ast \omega \).

For arbitrary \(M\) and \(N\) with \(\alpha _i\) and \(\omega _i\) given as above, define \(\mu _i \in \Omega ^1(M \times N)\) by

\[ \mu _i := \pi _1^\ast \alpha _i - \pi _2^\ast \omega _i \]

and let \(\mathcal{I}\) be the ideal generated by the \(\mu _i\).

This gives a strategy for solving Problem 3.8.13: suppose there were a map satisfying our condition \(f^\ast \omega _i = \alpha _i\) for all \(i\), and let \(\Gamma \colon \thinspace \! M \to M \times N\) be the graph of \(f\): \(\Gamma (x) = (x,f(x))\) for all \(x \in M\). Then I claim that \(\Gamma \) is an integral manifold of \(\mathcal{I}\) (really, of the distribution \(\mathcal{D}\) so that \(\mathcal{I} = \mathcal{I}(\mathcal{D})\)).

To see this, it suffices to show that \(\Gamma ^\ast \mu _i = 0\) for all \(i\). Since \(\pi _1 \circ \Gamma = \operatorname {id}_M\) and \(\pi _2 \circ \Gamma = f\),

\[ \Gamma ^\ast \mu _i = \Gamma ^\ast (\pi _1^\ast \alpha _i - \pi _2^\ast \omega _i) = \Gamma ^\ast (\pi _1^\ast f^\ast \omega _i - \pi _2^\ast \omega _i) = (\pi _1 \circ \Gamma )^\ast f^\ast \omega _i - (\pi _2 \circ \Gamma )^\ast \omega _i = f^\ast \omega _i - f^\ast \omega _i = 0. \]

So this says that if \(f\) exists, its graph is an integral manifold of the ideal \(\mathcal{I}\), which of course means that \(\mathcal{I}\) must be a differential ideal (by Theorem 3.8.8). In other words, \(\mathcal{I}\) being a differential ideal is a necessary condition for such an \(f\) to exist.

3.9 Lie Group Homorphisms are Determined by their Differentials

Let’s recap and see where we are. We’re working our way towards a proof of Theorem 3.6.1, which (essentially) says that Lie group homomorphisms are uniquely determined their differentials and that every Lie algebra homomorphism is the differential of some Lie group homomorphism.

To do that, we want to prove the intermediate result Theorem 3.8.12, in which we’re trying to show that if \(f_1^\ast , f_2^\ast \colon \thinspace \! \mathfrak {h}^\ast \to \mathfrak {g}^\ast \) have \(f_1^\ast \omega _i = f_2^\ast \omega _i\) for some basis \(\omega _1, \dots , \omega _n\) for left-invariant 1-forms on \(H\), then \(f_1 \equiv f_2\).

In turn, we saw this as a special case of Problem 3.8.13, in which we have manifolds \(M\) and \(N\), a basis \(\omega _1, \dots , \omega _n\) for the 1-forms on \(N\) (which doesn’t always exist, but does on any Lie group) and some preferred \(\alpha _1 , \dots \alpha _n \in \Omega ^1(M)\), and we would like to find a map \(f \colon \thinspace \! M \to N\) so that \(f^\ast \omega _i = \alpha _i\) for all \(i=1, \dots , n\). We’ve seen that the ideal \(\mathcal{I}\) on \(M \times N\) generated by the 1-forms \(\mu _1 = \pi _1^\ast \alpha _i - \pi _2^\ast \omega _i \in \Omega ^1(M \times N)\) must be a differential ideal if such an \(f\) exists. In fact, this is also a sufficient condition for the existence of such an \(f\), which is essentially unique:

In particular, I claim that \(\left.d\pi _1 \right|_S\) is injective. To see this, let \((r,s) \in S\) and let \(v \in T_{(r,s)} S\). If \(d\pi _1(v) = 0\), then

1. \(\mu _i(v) = 0\) for all \(i=1, \dots , n\) since \(\mathcal{I}\) consists of all the forms which vanish on the tangent space to \(S\).

2. \(\pi _1^\ast \alpha _i(v) = \alpha _i(d\pi _1 v) = \alpha _i(0) = 0\).

But then these two facts together imply that

for all \(i=1, \dots , n\). Since the \(\omega _i\) form a basis of 1-forms on \(N\), this means that \(d\pi _2(v) = 0\).

In turn, this implies that \(v = 0\) since \(T_{(r,s)} M \times N = T_r M \oplus T_s N\). Since \((r,s) \in S\) was arbitrary, we see that \(\left. d\pi _1 \right|_S\) is injective, so the Inverse Function Theorem (Theorem 1.4.7) implies that \(\left. \pi _1 \right|_S \colon \thinspace \! S \to M\) is a local diffeomorphism since \(\dim S = (m+n)-n=m=\dim M\). Therefore, there exists a neighborhood \(V\) of \((p,q)\) in \(S\) and a neighborhood \(U\) of \(p\) in \(M\) so that \(\left. \pi _1 \right|_V \colon \thinspace \! V \to U\) is a diffeomorphism.

We’re finally ready to define \(f \colon \thinspace \! U \to N\) by \(f = \pi _2 \circ \left(\left. \pi _1\right|_V\right)^{-1}\). Of course now we need to check that \(f^\ast \omega _i = \alpha _i\) for all \(i=1, \dots , n\). To see this, pick \(i \in \{ 1, \dots , n\} \) and \(u \in T_pM\). Then

by the definition of \(\mu _i = \pi _1^\ast \alpha _i - \pi _2^\ast \omega _i\). The second term above is zero since \(d\left(\left. \pi _1 \right|_V\right)^{-1} u \in T_{(r,s)}S\) and \(\mathcal{I} = \mathcal{I}(S)\) is generated by the \(\mu _i\). Hence, we can continue the above computation to see

since \(\pi _1 \circ \left(\left. \pi _1 \right|_V\right)^{-1}\) is just the identity map.

Now let’s apply this result to proving Theorem 3.8.12, which I restate for convenience:

Suppose \(\omega _1, \dots , \omega _n\) is a basis for the left-invariant 1-forms on \(H\). Then we know that \(f_1^\ast \omega _i = f_2^\ast \omega _i\) for all \(i=1, \dots , n\). By uniqueness in Theorem 3.9.1, it will follow that \(f_1 \equiv f_2\) if we can show that the ideal \(\mathfrak {I}\) on \(G \times H\) generated by the

is a differential ideal. It suffices to show that the \(d\mu _i \in \mathcal{I}\), since a general element of \(\mathcal{I}\) will be a linear combination of terms of the form \(\beta _1 \wedge \dots \wedge \beta _\ell \wedge \mu _i \wedge \beta _{\ell +1} \wedge \dots \wedge \beta _k\), the exterior derivative of which will be a linear combination of wedge products which always include either \(\mu _i\) or \(d\mu _i\).

So let’s compute. First, notice that \(\omega _1, \dots , \omega _n\) is a basis for the left-invariant forms on \(H\), so \(\left\{ \omega _i \wedge \omega _j : i {\lt} j \right\} \) is a basis for the left-invariant 2-forms on \(H\). Since \(d\omega _k\) is left-invariant for any \(k \in \{ 1, \dots , n\} \), there exist constants \(c_{ij}^k\) so that

But then

by adding and subtracting a term of the form \(\pi _2^\ast \omega _i \wedge \pi _1^\ast f^\ast \omega _j\) in going from the fourth to the fifth line. But then each \( \mu _i\wedge \pi _1^\ast f_1^\ast \omega _j + \pi _2^\ast \omega _i \wedge \mu _j \in \mathcal{I}\), so we conclude that \(d\mu _k \in \mathcal{I}\), and hence \(\mathcal{I}\) is a differential ideal.

With all of this done, we have now proved uniqueness in Theorem 3.6.1, which I restate:

I also restate Corollary 3.6.2, which follows immediately:

Since we have proved uniqueness in Theorem 3.6.1, it remains to show existence, which we do by defining \(\phi \colon \thinspace \! G \to H\) as in the proof of Theorem 3.9.1 by \(\phi = \pi _2 \circ \left(\left. \pi _1\right|_S\right)^{-1}\) with \(S \subset G \times H\) with \(\mathcal{I} = \mathcal{I}(S)\). The key point is that the definition of the \(\mu _i = \pi _1^\ast \psi ^\ast \omega _i - \pi _2^\ast \omega _i\) which generate \(\mathcal{I}\) depends only on \(\psi \colon \thinspace \! \mathfrak {g} \to \mathfrak {h}\) and not on the \(\phi \) that needs to be constructed. Proving that \(\phi \) is really a homomorphism depends on some covering space theory that I don’t really want to get into, but suffice it to say that this is true.

Notice that the simply-connected hypothesis in Corollary 3.6.2 is essential, since \(\operatorname {O}(n)\) and \(\operatorname {SO}(n)\) have identical Lie algebras even though they’re not isomorphic as Lie groups (nor even homeomorphic). Admittedly, this feels like a cheat since it’s really only using connectedness, and of course \(\operatorname {O}(n)\) and \(\operatorname {SO}(n)\) do have the same connected component of the identity, which is all that the Lie algebra could plausibly detect.

But we’ve already seen a less trivial example in Proposition 3.4.3, which showed that \(\mathfrak {so}(3) \cong \mathfrak {su}(2)\), yet \(\operatorname {SO}(3)\) and \(\operatorname {SU}(2)\) are not isomorphic as groups (nor even homeomorphic as topological spaces). It is fairly easy to see that these groups are not isomorphic since they have different centers:

In fact, as topological spaces \(\operatorname {SO}(3) \cong \mathbb {RP}^3\) and \(\operatorname {SU}(2) \cong S^3\), so these spaces have different fundamental groups: \(\pi _1\left(\mathbb {RP}^3\right) \cong \mathbb {Z}/2\mathbb {Z}\) and \(\pi _1\left(S^3\right) = \{ 1\} \). Hence, they cannot be homeomorphic either. Looked at this way, \(S^3\) is the universal cover of \(\mathbb {RP}^3\), meaning that \(\operatorname {SU}(2)\) is homeomorphic to the universal cover of \(\operatorname {SO}(3)\), and hence also to \(\operatorname {Spin}(3)\). ³¹

3.10 The Exponential Map

In classifying Lie groups (especially compact Lie groups), it turns out to be particularly helpful to understand the abelian subgroups, and especially the maximal abelian subgroups (so-called maximal tori). The simplest abelian subgroups are called one-parameter subgroups:

This is a general phenomenon: if \(G\) is a Lie group and \(X \in \mathfrak {g}\), then the map \(\lambda \frac{d}{dr} \mapsto \lambda X\) defines a homomorphism from the Lie algebra of \(\mathbb {R}\) into \(\mathfrak {g}\). Since \(\mathbb {R}\) is simply-connected, Theorem 3.6.1 guarantees there exists a unique one-parameter subgroup \(\phi _X \colon \thinspace \! \mathbb {R}\to G\) so that \(d\phi _X\left(\lambda \frac{d}{dr} \right) = \lambda X\).

Example 3.10.5

Let \(G = \operatorname {U}(1)\), the unit complex numbers. We can identify \(\mathfrak {u}(1)\) with the tangent space at the identity \(T_1 \operatorname {U}(1)\), which is visually just a copy of the imaginary axis, as we see in Figure 3.8.

We can make the identification \(\mathfrak {u}(1) \cong i \mathbb {R}\). Let \(X = i a \in \mathfrak {u}(1)\) and define the map \(\phi _X \colon \thinspace \! \mathbb {R}\to \operatorname {U}(1)\) by \(\phi _X(r) = e^{rX} = e^{i a r}\). Since \(\phi _X'(0) = ia = X\), this is the one-parameter subgroup guaranteed by Theorem 3.6.1, and so by definition

\[ \exp (X) = \phi _X(1) = e^X = e^{ia}. \]

This helps explain the source of the name: the exponential map on \(\operatorname {U}(1)\) is just the usual exponential map from complex analysis.

More generally, suppose \(A \in \operatorname {Mat}_{n\times n}(\mathbb {R}) \cong \mathfrak {gl}_n(\mathbb {R})\) and define \(\gamma (t) = e^{tA}\), where the matrix exponential is defined by the Taylor series

(which turns out to converge everywhere, just like the usual Taylor series for the single variable exponential). Then

so \(\gamma (t) = \phi _X(t)\) and hence \(\exp (A) = \gamma (1) = e^A\). In other words, the exponential map on matrix groups is always just the matrix exponential.

Some straightforward properties of the exponential:

In general, the exponential map is how we translate between Lie algebra homomorphisms and Lie group homomorphisms:

Theorem 3.10.10

Let \(\phi \colon \thinspace \! H \to G\) be a Lie group homomorphism. Then the following diagram commutes:

Moreover, it’s a one-parameter subgroup since \(G\) is a homomorphism, so by uniqueness

for all \(t\), including \(t=1\), which gives \(\phi (\exp (X)) = \gamma (1) = \exp (d\phi (X))\), as desired.

3.11 Homogeneous Manifolds

Given a group \(G\) and a normal subgroup \(N\), it’s a standard fact from algebra that the quotient (i.e., collection of cosets) \(G/N\) is also a group. The key point is that, for \(g_1, g_2 \in G\), the coset \((g_1 N) (g_2 N) := g_1 g_2 N\) is well-defined: for \(n_1,n_2 \in N\), we can write \(n_1 = g_2 n g_2^{-1}\) for \(n := g_2^{-1} n_1 g_2 \in N\), so

The same holds true for Lie groups: if \(G\) is a Lie group and \(N\) is a normal Lie subgroup, then \(G/N\) is a Lie group. For example, \(\operatorname {SL}_n(\mathbb {R})\) is a closed normal subgroup of \(\operatorname {GL}_n(\mathbb {R})\), and

the multiplicative group of nonzero real numbers

Okay, but what if we take a Lie subgroup \(H\) of a Lie group \(G\) which is not normal? Does the coset space \(G/H\) have any special structure? Not surprisingly, given how I’ve set it up, the answer is yes:

The proof of this theorem is kind of a pain, so we skip it. Manifolds of the form \(G/H\) are called homogeneous manifolds (or sometimes homogeneous spaces).

Essentially the same argument gives a much more general result, which we need one definition in order to state:

(Incidentally, this induces the isotropy representation \(G_p \to \operatorname {Aut}(T_pM)\) given by \(g \mapsto \left.(d \phi _g)_p \right|_{T_pM}\), where \(\phi _g(q) := g \cdot q\) for all \(q \in M\).)

Indeed, most books define homogeneous manifolds to be manifolds admitting a smooth, transitive action of a Lie group, which explains the name: since the group action is transitive, every point is in some sense the same as every other point. Theorem 3.11.4 tells us that our definition of homogeneous manifold and the standard definition are equivalent.

3.12 Stiefel and Grassmann Manifolds

Stiefel manifolds and Grassmann manifolds (or Grassmannians) are very important classes of homogeneous manifolds which show up all over the place in both pure and applied mathematics. In turn, the simplest example of a Grassmannian is a projective space:

Example 3.12.1

Complex projective space \(\mathbb {CP}^n\) is the collection of all 1-dimensional subspaces of \(\mathbb {C}^{n+1}\); equivalently,

\[ \mathbb {CP}^n = \left\{ (x_0, \dots , x_n) \in \mathbb {C}^{n+1} - \{ 0\} \right\} /\sim , \]

where \((x_0, \dots , x_n) \sim \lambda (x_0, \dots , x_n)\) for all \(\lambda \in \mathbb {C}^\times \).

If we restrict to unit vectors, we can also see

\[ \mathbb {CP}^n = \left\{ (x_0, \dots , x_n) \in S^{2n+1} \subset \mathbb {C}^{n+1}\right\} /\sim , \]

where now \((x_0, \dots , x_n) \sim \lambda (x_0, \dots , x_n)\) for all \(\lambda \in U(1)\).

We typically represent the equivalence class of a point \((x_0, \dots , x_n)\) by homogeneous coordinates \([x_0 : \dots : x_n]\).

As we saw in Example 3.11.7, \(\operatorname {SU}(n+1)\) acts transitively on \(S^{2n+1}\). Moreover, this action preserves equivalence classes since

\[ A(\lambda (x_0, \dots , x_n)) = \lambda A(x_0, \dots , x_n) \sim A(x_0, \dots , x_n) \]

by complex-linearity of \(A\), so \(\operatorname {SU}(n+1)\) acts transitively on \(\mathbb {CP}^n\). The isotropy subgroup of \([1:0: \dots : 0] \in \mathbb {CP}^n\) is just

\[ \left\{ \begin{bmatrix} \frac{1}{\det A} & 0 \\ 0 & A \end{bmatrix} : A \in \operatorname {U}(n)\right\} \cong \operatorname {U}(n), \]

so \(\mathbb {CP}^n \cong \operatorname {SU}(n+1)/\operatorname {U}(n)\) is a homogeneous space.

Moreover, since \(\operatorname {SU}(n+1) \cong \operatorname {U}(n)/\operatorname {U}(1)\) we have that

$$\begin{multline*} \mathbb {CP}^n \cong \operatorname {SU}(n+1)/\operatorname {U}(n) \cong \left(\operatorname {U}(n+1)/\operatorname {U}(1)\right)/\operatorname {U}(n) \cong \operatorname {U}(n+1)/\left(\operatorname {U}(1) \times \operatorname {U}(n)\right) \\ \cong \left(\operatorname {U}(n+1)/\operatorname {U}(n)\right)/\operatorname {U}(1) \cong S^{2n+1}/\operatorname {U}(1) \end{multline*}$$

by Example 3.11.6. Therefore, we have a fibration

Another interpretation: the intersection of each complex line with \(S^{2n+1}\) is a circle of the form \(\lambda v\) where \(\lambda \) is a unit complex number and \(v \in S^{2n+1}\). So then the equivalence classes of such circles exactly correspond to complex lines in \(\mathbb {C}^{n+1}\).

Notice that the fiber is a sphere: \(\operatorname {U}(1) \cong S^1\).

In the special case \(n=1\), the resulting fibration is

usually called the Hopf fibration; here’s a nice Hopf fibration video by Niles Johnson.

Based on this, the fibrations for \(n {\gt} 1\) are sometimes called generalized Hopf fibrations. We can play similar games with \(\mathbb {HP}^n\) and \(\mathbb {OP}^1\) to get other sphere fibrations.

While the above is an example of a Grassmannian, let’s define Stiefel manifolds before defining Grassmannians:

In other words, elements of \(\operatorname {St}(k,\mathbb {R}^n)\) are \(k\)-tuples of pairwise orthonormal vectors in \(\mathbb {R}^n\) (with respect to, say, the standard inner product). Of course this implies that \(v_1, \dots , v_k\) are linearly independent, so \(k \leq n\).

\(\operatorname {O}(n)\) acts transitively on \(\operatorname {St}(k,\mathbb {R}^n)\), which we can see because any orthogonal matrix with \((v_1, \dots , v_k)\) as its first \(k\) columns sends the first \(k\) standard basis vectors \((e_1, \dots , e_k)\) (which is a particular point in the Stiefel manifold) to \((v_1, \dots , v_k)\). Therefore, \(\operatorname {St}(k,\mathbb {R}^n) \cong \operatorname {O}(n)/H\) where \(H \subset \operatorname {O}(n)\) is an isotropy subgroup. Consider the point \((e_1, \dots , e_k) \in \operatorname {St}(k, \mathbb {R}^n)\). Then the isotropy subgroup of this point is

so \(\operatorname {St}(k,\mathbb {R}^n) \cong \operatorname {O}(n)/\operatorname {O}(n-k)\). Notice that this agrees with \(S^{n-1} \cong \operatorname {O}(n)/\operatorname {O}(n-1)\) when \(k=1\) and with \(\operatorname {O}(n) \cong \operatorname {O}(n)/\operatorname {O}(0) = \operatorname {O}(n)/\{ I\} \) when \(k=n\).

In particular, this implies that

As in the case of the Stiefel manifold, \(\operatorname {O}(n)\) acts transitively on \(\operatorname {Gr}(k,\mathbb {R}^n)\), and the isotropy subgroup of the \(k\)-plane \(\operatorname {span}\{ e_1, \dots , e_k\} \) is

so \(\operatorname {Gr}(k,\mathbb {R}^n) \cong \operatorname {O}(n)/(\operatorname {O}(k) \times \operatorname {O}(n-k))\).

In particular, this shows that

Taking quotients in stages reveals that

Here’s a more geometric interpretation of this: given \(P \in \operatorname {Gr}(k,\mathbb {R}^n)\), one could find an orthonormal basis \(v_1, \dots , v_k\) for \(P = \operatorname {span}\{ v_1,\dots , v_k\} \); then \((v_1, \dots , v_k) \in \operatorname {St}(k,\mathbb {R}^n)\) and the map \((v_1, \dots , v_k) \mapsto \operatorname {span}\{ v_1, \dots , v_k\} \) defines a map \(\operatorname {St}(k,\mathbb {R}^n) \to \operatorname {Gr}(k,\mathbb {R}^n)\). Two elements of the Stiefelm manifold get sent to the same element of the Grassmannian if they are two different orthonormal bases for the same \(k\)-plane, which means they must be related by an orthogonal transformation of the \(k\)-plane. Identifying \((v_1, \dots , v_k)\) with the (tall, skinny) matrix whose columns are the \(v_i\), this corresponds to right-multiplying by an element of \(\operatorname {O}(k)\), so indeed elements of \(\operatorname {Gr}(k,\mathbb {R}^n)\) correspond to (right) \(\operatorname {O}(k)\)-orbits of points in \(\operatorname {St}(k,\mathbb {R}^n)\).

This perspective naturally leads to the so-called Stiefel bundles (or, in fancy language, the frame bundle associated to the tautological bundle of the Grassmannian):

In general, if \(V\) is a vector space, we can define the Grassmannian \(\operatorname {Gr}(k,V)\) to be the space of \(k\)-dimensional linear subspaces of \(V\). Also, when \(V\) is an inner product space (e.g., \(\mathbb {R}^n\) or \(\mathbb {C}^n\)), we can define the Stiefel manifold \(\operatorname {St}(k,V)\) of \(k\)-tuples of pairwise orthonormal vectors in \(V\).

When \(V = \mathbb {C}^n\), it is straightforward to show that \(\operatorname {St}(k,\mathbb {C}^n) \cong \operatorname {U}(n)/\operatorname {U}(n-k)\) and \(\operatorname {Gr}(k,\mathbb {C}^n) \cong \operatorname {U}(n)/(\operatorname {U}(k) \times \operatorname {U}(n-k))\). We get a corresponding complex Stiefel bundle:

When \(k=1\) this is just the generalized Hopf fibration from Example 3.12.1, so these Stiefel bundles are in some sense generalized generalized Hopf fibrations.

I will also briefly mention that Grassmannians are special cases of more general homogeneous manifolds called flag manifolds (or flag varieties). In short, for \(1 \leq d_1 {\lt} d_2 {\lt} \dots {\lt} d_{k-1} {\lt} d_k = n\), a flag of signature \((d_1, \dots , d_k)\) is a nested collection of subspaces of \(\mathbb {R}^n\) (or \(\mathbb {C}^n\))

The collection of all flags of signature \((d_1, \dots , d_k)\) is the flag manifold \(\operatorname {F\ell }(d_1, \dots , d_k)\). Notice that a flag of signature \((d,n)\) is just a single \(d\)-dimensional subspace \(V \subset \mathbb {R}^n\); that is, a point in the Grassmannian \(\operatorname {Gr}(d,\mathbb {R}^n)\). Therefore, \(\operatorname {F\ell }(d,n) = \operatorname {Gr}(d,\mathbb {R}^n)\), so the flag manifolds really are generalizations of Grassmannians.

At the other end of the spectrum, flags of signature \((1,2,\dots , n-1,n)\) are called complete flags, and \(\operatorname {F\ell }(1,2,\dots , n)\) is called the complete flag manifold (or complete flag variety). ³²

The coadjoint orbits of the unitary group \(\operatorname {U}(n)\) turn out to give all of the complex flag manifolds consisting of flags in \(\mathbb {C}^n\). For more on this, see, e.g., Audin’s book  [ 2 , § II.1.d ] .

4.1 Riemannian Metrics

Thus far, almost all of what we’ve talked about is, strictly speaking, in the realm of differential topology rather than differential geometry. The exceptions include Section 2.2, Example 2.3.22, and some of the distribution theory (including contact geometry) from Section 3.7 and Section 3.8. The point is that, to have geometry we need some notion of size, whether that’s in the form of length, area, volume, or something else.

A weak form of this is provided by volume forms, and a somewhat stronger form by symplectic forms, which encode a notion of area (and hence volume), but not of length. In this section we introduce Riemannian metrics, which induce a notion of length (and hence also of area, volume, etc.). We’ve already given a definition in Definition 2.3.21, which we repeat:

Let \((U,\phi )\) be some local coordinate chart on \(M\) and let \(\frac{\partial }{\partial x_1}, \dots , \frac{\partial }{\partial x_n}\) be the induced local coordinate basis on the tangent spaces of points in the image of \(\phi \). Then we can represent \(g\) by the symmetric matrix \(\begin{bmatrix} g_{ij} \end{bmatrix}_{i,j}\) where

In the above I’ve written \(g_{ij}\) as a function on \(\phi (U) \subset M\), but it is also common to think of \(g_{ij}\) as a function on \(U \subset \mathbb {R}^n\): \(g_{ij}(x_1, \dots , x_n) = g_{ij}(\phi (x_1, \dots , x_n))\). Of course, the fact that \(g\) is a smooth tensor field implies that, in this interpretation, \(g_{ij}\) is a smooth function on \(U\). ³³

Before giving examples, let’s define the appropriate notion of isomorphism of Riemannian metrics.

I promised above that Riemannian metrics give a notion of length on manifolds. Of course, given a Riemannian metric, I get a notion of lengths of tangent vectors by just taking the norm associated to the inner product: for \(v \in T_p M\), its norm/length is just \(\sqrt{g_p(v,v)}\).

But a Riemannian metric also gives a notion of lengths of curves by integrating the norm of the tangent vector to the curve:

Given a notion of length, we also get a notion of volume. First, if we choose some orthonormal basis \(e_1, \dots , e_n\) for \(T_p M\) and consider the coordinate vectors \(X_i(p) := \frac{\partial }{\partial x_i}(p)\), then there exist \(a_{ij}\) so that

The volume \(\operatorname {vol}(X_1(p), \dots , X_n(p))\) of the parallelpiped spanned by the \(X_i\) in \(T_pM\) is equal to \(\operatorname {vol}(e_1, \dots , e_n) = 1\) times the determinant of the change-of-basis matrix \(A := \begin{bmatrix} a_{ij} \end{bmatrix}_{i,j}\):

To relate this to the metric, observe that

which is just the \((i,j)\) entry of \(AA^T\). Since

we can combine (4.1) and (4.2) to see that

where \(\sqrt{\det g_{ij}}\) is the standard shorthand for \(\sqrt{\det \begin{bmatrix} g_{ij} \end{bmatrix}_{i,j}}\).

Notice that, if \(p\) lies in another local coordinate chart \((V,\psi )\) with coordinate basis \(Y_i(p) = \frac{\partial }{\partial y_i}(p)\), then in those coordinates the Riemannian metric looks like \(\begin{bmatrix} h_{ij} \end{bmatrix}_{i,j}\) with \(h_{ij}(p) := g_p(Y_i(p), Y_j(p))\) and

where \(J = \begin{bmatrix} \frac{\partial x_i}{\partial y_j} \end{bmatrix}_{i,j}\) is the Jacobian of the change of coordinates map \(\psi ^{-1} \circ \phi \colon \thinspace \! U \to V\).

This is well-defined: if \(R \subset \psi (V)\) for some other local coordinate chart \((V,\phi )\), then

by (4.3) and the change of variables formula.

More generally, we can define the volume of a region not contained in a single coordinate chart using a partition of unity, as in Definition 2.9.6. Indeed, the point is really that a Riemannian metric determines a volume form:

If \(\operatorname {vol}_g(M) = \int _M \operatorname {dVol}_g\) is finite (e.g., if \(M\) is compact), then the Riemannian volume form determines a (signed) measure, called the Riemannian volume measure, on the Borel sets in \(M\):

(If you don’t want a signed measure, just take the absolute value of the right hand side.)

So, for example, if \((M,g)\) is a compact Riemannian manifold, then we automatically get a probability measure on \(M\) by normalizing the above measure:

4.2 Affine Connections

We would like to define geodesics on Riemannian manifolds. These are the analogs of straight lines, in the sense both that they are defined as “straightest” paths on Riemannian manifolds and that shortest paths between points are always geodesics. To do so, we will need to define an intrinsic notion of the derivative of a vector field along a curve; then the geodesics will be the curves whose tangent vectors differentiate to zero (just like the tangent vector to a straight line parametrized at constant speed is constant, and hence its derivative is zero; or, if you like, the acceleration along a straight line is zero).

This notion of differentiation will be called the covariant derivative. If our manifold is a smooth surface in \(\mathbb {R}^3\) (with the Riemannian metric induced by the Euclidean structure on \(\mathbb {R}^3\)), then there’s a simple geometric description of the covariant derivative, which I will now try to explain.

Let \(S\) be a smooth surface in \(\mathbb {R}^3\) and let \(\alpha \colon \thinspace \! I \to S\) be a smooth curve in \(S\), where \(I\) is some interval. Suppose \(V\) is a vector field along \(\alpha \). Concretely, we can write \(V \colon \thinspace \! I \to \mathbb {R}^3\) so that \(V(t) \in T_{\alpha (t)}S\): that is, at each \(t\), the vector \(V(t)\) is tangent to the curve \(\alpha \).

In general, there is no reason that \(V'(t)\) should be tangent to \(S\).

Example 4.2.1

Consider the case where \(S\) is the unit sphere, \(\alpha (t) = \left(\frac{1}{\sqrt{2}}\cos t, \frac{1}{\sqrt{2}}\sin t, \frac{1}{\sqrt{2}} \right)\) is the circle of latitude at \(45^\circ \) North, and

\[ V(t) := \left( \sin ^2 t - \frac{1}{\sqrt{2}} \cos ^2 t, -\frac{2+\sqrt{2}}{4} \sin 2t, \frac{1}{\sqrt{2}} \cos t\right). \]

See Figure 4.1.

Then

\[ V'(t) = \left((2 + \sqrt{2})\cos t \sin t,- \frac{2 + \sqrt{2}}{2}\cos 2t, - \frac{\sin t}{\sqrt{2}}\right). \]

Hence \(V'(\pi /2) = \left(0,1+\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)\) is not in the tangent space to \(\alpha (\pi /2) = \left(0, \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\) since

\[ V'(\pi /2) \cdot \alpha (\pi /2) = \frac{1}{\sqrt{2}} \neq 0. \]

The fact that \(V'(t)\) is not generally tangent to \(S\) means that \(V'(t)\) is not a notion of derivative that can be defined intrinsically to the surface. However, in this case, the simplest possible fix turns out to give something that is intrinsically defined. Specifically, we know that \(\alpha (t) \in S\), and so we have the tangent space \(T_{\alpha (t)}S\) to \(S\) at \(\alpha (t)\). Of course, \(\alpha (t) \in \mathbb {R}^3\) as well, and \(\mathbb {R}^3\) has its own tangent space at \(\alpha (t)\), namely \(T_{\alpha (t)}\mathbb {R}^3\). But then \(V'(t) \in T_{\alpha (t)}\mathbb {R}^3\) and \(T_{\alpha (t)} S \subset T_{\alpha (t)}\mathbb {R}^3\), so we can orthogonally project \(V'(t)\) to \(T_{\alpha (t)}S\), as in Figure 4.2.

The projection of \(V'(t)\) onto \(T_{\alpha (t)}S\) is called the covariant derivative of \(V\) along \(\alpha \) and denoted \(\frac{DV}{dt}\). While this still seems extrinsic to the surface, it turns out you can define \(\frac{DV}{dt}\) purely in terms of \(\alpha \), \(V\), \(S\), and the Riemannian metric \(g\) on \(S\). While in this example the Riemannian metric is induced by the Euclidean metric on \(\mathbb {R}^3\), you can use the same formula to define \(\frac{DV}{dt}\) for any Riemannian metric \(g\) on a surface, regardless of whether or not \(g\) is induced from a Euclidean metric.

In turn, we say that a vector field \(V\) along \(\alpha \) is parallel if \(\frac{DV}{dt} \equiv 0\); then a geodesic is just a curve \(\alpha \) so that \(\alpha '(t)\) is parallel along \(\alpha \).

If you want to see more details on the above, see any standard textbook on curves and surfaces, for example do Carmo’s book  [ 8 ] or O’Neill’s  [ 26 ] . Here, I’m going to develop the general theory on arbitrary Riemannian manifolds, which uses a gadget called a connection. These can be defined on any manifold (not just Riemannian manifolds), but, as we’ll see, a Riemannian metric produces a special connection.

by 3. Since the support of the cutoff function \(g\) can be arbitrarily small, we see that \(\nabla _V W\) depends only on the values of \(W\) in a small neighborhood of \(p\).

From 1, we know \(\nabla _{fV}W = f \nabla _V W\), so it follows immediately that \(\nabla _V W\) at \(p\) also depends only on the values of \(V\) in a neighborhood of \(p\).

In fact, we can show more. Let \((U,\phi )\) be a coordinate chart in a neighborhood of \(p\) and let \(X_1 = \frac{\partial }{\partial X_1}, \dots , X_n = \frac{\partial }{\partial x_n}\) be the coordinate vector fields. If \(V = \sum _i v_i X_i\), then

by 1. If we evaluate the right hand side at \(p\), the only dependence on \(V\) come from the \(v_i(p)\), so it only depends on \(V(p)\) (and not on the behavior of \(V\) in a neighborhood of \(p\)).

If \(\nabla \) is an affine connection on \(M\) and \(X_1 = \frac{\partial }{\partial X_1}, \dots , X_n = \frac{\partial }{\partial x_n}\) in some coordinate neighborhood on \(M\), define the smooth functions \(\Gamma _{ij}^k\) on that coordinate neighborhood by

The \(\Gamma _{ij}^k\) are called the Christoffel symbols of the connection \(\nabla \), and we can write a formula for \(\nabla _V W\) in terms of them: if \(V = \sum _i v_i X_i\) and \(W = \sum _j w_j X_j\) are vector fields defined on a coordinate neighborhood, then

We can refine Lemma 4.2.4 a bit: we don’t actually need to know \(W\) on an entire neighborhood of \(p\), but only along a curve within that neighborhood:

since \(V = \sum _i v_i \frac{\partial }{\partial x_i}\). The only term which is not entirely determined by \(V(p)\) and \(W(p)\) is \(V(w_k) = (\alpha \circ w_k)'(0)\) for some curve \(\alpha \) with \(\alpha (0) = p\) and \(\alpha '(0) = V(p)\). Of course, \((\alpha \circ w_k)(t)\) only depends on the values of \(W\) along \(\alpha \).

An affine connection then gives us a way of defining covariant derivatives:

so the formula (4.4) for \(\nabla _VW\) shows that we must define

in order to satisfy conditions 1, 2, and 3. Hence, the covariant derivative, if it exists, must be unique. But to get existence we just define the covariant derivative in coordinates as in (4.5).

The thing to remember about affine connections is that they give a way of defining parallel transport; that is, some way of moving a tangent vector “parallel to itself” along a curve. In general, there’s no canonical way to do this, so you have to define what you mean by “parallel,” and that’s really what a connection does.

Example 4.2.10

Consider the unit sphere \(S^2 \subset \mathbb {R}^3\) and take the tangent vector \(\frac{\partial }{\partial y}\) at the north pole. Say we want to transport this vector parallel to itself around the spherical triangle that traverses a great circle south from \((0,0,1)\) to \((1,0,0)\), then the equator east from \((1,0,0)\) to \((0,1,0)\), and finally another great circle north from \((0,1,0)\) to \((0,0,1)\). If course, we could move \(\frac{\partial }{\partial y}\) parallel to itself in \(\mathbb {R}^3\), which just gives \(\frac{\partial }{\partial y}\) at every point along the curve; see Figure 4.3. But \(\frac{\partial }{\partial y}\) is not in the tangent space at \((0,1,0) \in S^2\): it’s perpendicular to the sphere at that point!

To get something which stays tangent to the sphere, we simply take the vector field which is always pointing east. Along the first segment, traveling south from the north pole to \((1,0,0)\), this is the same thing we had before. You can imagine a traveler walking south along this path, and then the vector field is just what they get when they stick out their left hand.

At \((1,0,0)\), the traveler turns left, and now east is straight ahead as they start walking toward \((0,1,0)\), so now the vector field is what they get by pointing straight ahead. From the perspective of the traveler, this is the same direction as along the first leg of their journey (namely, east!).

When they arrive at \((0,1,0)\), the traveler now turns left again to face north. Now east is to their right, and they can trace out the vector field by just sticking out their right hand as they walk north. Notice that, once the traveler gets back to the north pole, the vector we get by parallel translating the initial vector around the curve is perpendicular to the initial vector! So even though all along the way the vector field was constant (from the perspective of the traveler or, more mathematically, with respect to the intrinsic geometry of the sphere), it realizes some nontrivial transformation of the initial vector. ³⁴ See Figure 4.4.

The idea of parallel translation is the same in general: the idea is that, as you traverse a curve, you’re not changing the direction of the vector.

It turns out that, given any initial vector at a point and any curve through the point, we can always find a parallel field along the curve which agrees with the given vector at the point.

From here, we could define geodesics, curvatures, etc. for arbitrary connections. Rather than develop this theory in full generality, we will specialize to the case of the Riemannian connection or Levi-Civita connection, which is uniquely determined by a Riemannian metric. So we turn to the task of defining this special connection.

4.3 The Levi-Civita Connection

We need to define two notions, compatibility and symmetry, before we can give the main theorem of this section, which is the existence of the Levi-Civita connection.

where the second line follows from Exercise 4.3.4 and the third from Proposition 4.2.83. Since \(p\) was chosen arbitrarily, (4.6) follows.

On the other hand, if \(\nabla \) satisfies (4.6) and \(V,W\) are parallel along \(\alpha \colon \thinspace \! I \to M\), then Exercise 4.3.4 implies that

since \(\frac{DV}{dt} \equiv 0\) and \(\frac{DW}{dt} \equiv 0\).

We are now ready to state and prove the main result:

This connection is usually called the Levi-Civita connection or the Riemannian connection.

Since the connection is symmetric, we know that

and similarly for other terms, so subtracting (4.9) from the sum of (4.7) and (4.8) yields

In other words,

which shows that \(\nabla \) is uniquely determined by \(g\). Hence, if it exists, \(\nabla \) must be unique. ³⁵

To prove existence, we can simply define \(\nabla \) by Equation 4.10; it is straightforward to check that \(\nabla \) defined in this way is a symmetric, compatible connection.

Example 4.3.10

Let’s work out the Levi-Civita connection for the Riemannian metric on \(\operatorname {Aff}^+(\mathbb {R})\) that we defined in Example 4.1.7. Recall that we can identify \(\operatorname {Aff}^+(\mathbb {R})\) with the upper half-plane \(H = \{ (x,y) \in \mathbb {R}^2 : y {\gt} 0\} \) and that the metric \(g\) can be represented by the matrix

$$\begin{align*} g_{(x,y)} \leftrightarrow \begin{bmatrix} g_{11}(x,y) & g_{12}(x,y) \\ g_{12}(x,y) & g_{22}(x,y) \end{bmatrix} = \begin{bmatrix} \frac{1}{y^2} & 0 \\ 0 & \frac{1}{y^2} \end{bmatrix}. \end{align*}$$

Let \(X = \frac{\partial }{\partial x}\) and \(Y = \frac{\partial }{\partial y}\). Since these are coordinate fields, we know \([X,Y] = 0\), and we always have \([X,X] = 0\) and \([Y,Y] = 0\) for any vector fields. So half of the terms in (4.10) vanish.

The matrix for \(g\) is telling us that

\[ g_{(x,y)}(X,X) = \frac{1}{y^2}, \quad g_{(x,y)}(X,Y) = 0, \quad g_{(x,y)}(Y,Y) = \frac{1}{y^2}, \]

and hence that, using (4.10)

$$\begin{align*} g_{(x,y)}(\nabla _X X,X) & = \frac{1}{2}\left[ X(g_{(x,y)}(X,X)) + X(g_{(x,y)}(X,X)) - X(g_{(x,y)}(X,X))\right] \\ & = \frac{1}{2}X(g_{(x,y)}(X,X)) \\ & = \frac{1}{2} \frac{\partial }{\partial x}\left(\frac{1}{y^2}\right) \\ & = 0 \end{align*}$$

and

$$\begin{align*} g_{(x,y)}(\nabla _X X,Y) & = \frac{1}{2}\left[ X(g_{(x,y)}(X,Y)) + X(g_{(x,y)}(Y,X)) - Y(g_{(x,y)}(X,X))\right] \\ & = \frac{1}{2}\left[X(0) + X(0) - Y\left(\frac{1}{y^2}\right)\right] \\ & = -\frac{1}{2} \frac{\partial }{\partial y}\left(\frac{1}{y^2}\right) \\ & = \frac{1}{y^3}. \end{align*}$$

Therefore, for any \(a\) and \(b\),

$$\begin{equation}\label{eq:affconn1} g_{(x,y)}(\nabla _X X,aX+bY) = \frac{b}{y^3}.\tag{4.11}\end{equation}$$

On the other hand, \(\nabla _X X = cX+dY\) for some \(c\) and \(d\), and we know that

$$\begin{equation}\label{eq:affconn2} g_{(x,y)}(cX+dY,aX+bY) = \frac{ac}{y^3} + \frac{bd}{y^3}.\tag{4.12}\end{equation}$$

Equating (4.11) and (4.12), we see that \(\nabla _XX = \frac{1}{y}Y\).

If we play the same game with \(\nabla _YY\) and with \(\nabla _XY = \nabla _YX\) (using the symmetry of \(\nabla \) and the fact that \([X,Y] = 0\)), we conclude that

$$\begin{equation}\label{eq:hyperbolic plane Christoffel} \nabla _XX = \frac{1}{y}Y , \quad \nabla _XY = \nabla _YX = -\frac{1}{y}X, \quad \nabla _YY = -\frac{1}{y}Y.\tag{4.13}\end{equation}$$

Now, let’s think about parallel transporting a vector along the curve \(\alpha (t) = (t,1)\). Notice that \(\alpha '(t) = X\) for all \(t\), so by Proposition 4.2.83, we have that the covariant derivative along \(\alpha \) is given by

\[ \frac{DW}{dt} = \nabla _{\alpha '(t)} W = \nabla _X W \]

for any tangent vector \(W\). So if we think of a vector field \(W(t) = a(t)X + b(t)Y \in T_{\alpha (t)}H = T_{(t,1)}H\), then the parallel transport equation becomes

$$\begin{multline*} 0 = \frac{DW}{dt} = \nabla _X (a(t)X + b(t)Y) = a'(t) X + a(t) \nabla _X X + b'(t)Y + b(t) \nabla _X Y \\ = a'(t) X +a(t)Y + b'(t) Y - b(t)X = (a'(t)-b(t))X + (b'(t)+a(t))Y \end{multline*}$$

since \(y=1\) along the whole curve and hence \(\frac{1}{y} = 1\). Of course, the coefficients have to be 0, so we get the homogeneous system of ODEs

$$\begin{align*} a'(t)-b(t) & = 0 \\ b'(t)+a(t) & = 0. \end{align*}$$

Solutions are of the form

$$\begin{align*} a(t) & = C_1 \cos t + C_2 \sin t \\ b(t) & = -C_1 \sin t + C_2 \cos t. \end{align*}$$

Figure 4.5 shows the parallel transports of \(X \in T_{(0,1)}H\) and \(Y \in T_{(0,1)}H\) along this curve.

  

Recall that a parallel vector field is always pointing in the same direction from the perspective of someone inside the manifold, so the way to interpret this is that a person traversing the curve \(\alpha (t)\) is constantly turning to the left to stay on \(\alpha \).

On the other hand, if \(\beta (t) = (0,t)\), then \(\beta '(t) = Y\) for all \(t\), and the parallel transport equation is

$$\begin{multline*} 0 = \frac{DW}{dt} = \nabla _Y (a(t)X + b(t)Y) = a'(t) X + a(t) \nabla _Y X + b'(t)Y + b(t) \nabla _Y Y \\ = a'(t) X -\frac{a(t)}{t}X + b'(t) Y - \frac{b(t)}{t}Y = \left(a'(t)-\frac{a(t)}{t})\right)X \left(b'(t)-\frac{b(t)}{t}\right)Y \end{multline*}$$

since \(y = t\) along the curve, and hence \(\frac{1}{y} = \frac{1}{t}\). So the system of ODEs is

$$\begin{align*} a'(t)-\frac{a(t)}{t} & = 0 \\ b'(t)-\frac{b(t)}{t} & = 0. \end{align*}$$

Solutions are of the form

$$\begin{align*} a(t) & = C_1t \\ b(t) & = C_2t. \end{align*}$$

Figure 4.6 shows the parallel transports of \(X \in T_{(0,1)}H\) and \(Y \in T_{(0,1)}H\) along this curve.

  

Notice that the parallel transport of a tangent vector to \(\beta \) stays tangent to \(\beta \); this is an indication that (up to reparametrization) \(\beta \) is a geodesic.

4.4 Geodesics

Geodesics are generalizations of straight lines to Riemannian manifolds. ³⁶ Typically we think of straight lines in the plane or in space as shortest paths, but this is actually the wrong notion to generalize. Not because shortest paths won’t be geodesics, but because not all geodesics are shortest paths.

Instead, the right notion to generalize is that straight lines are (no surprise) straightest paths. We can see this in terms of formulas by taking a constant-speed parametrization of a straight line, something like \(\alpha (t) = p + tv\), for \(p,v \in \mathbb {R}^n\). Then \(\alpha '(t) = v\) is constant, and hence \(\alpha ''(t) = 0\). The equation \(\alpha ''(t) = 0\) is going to generalize to the so-called geodesic equation.

Let’s work out what the geodesic equation (4.14) looks like in local coordinates. So work in a local coordinate chart on \(M\). Suppose \(\gamma (t) = (x_1(t),\dots , x_n(t))\) is a curve and \(W(t) = \sum _j w_j(t) \frac{\partial }{\partial x_j}\) is a vector field along \(\gamma \). We saw in (4.4) that

So if

then we have

Then \(\gamma \) is a geodesic if and only if the above is zero, which means all the coefficients must be zero:

which is a system of \(n\) second-order ODEs on a coordinate neighborhood of \(M\).

Example 4.4.4

Let’s figure out the geodesic equation on \(\operatorname {Aff}^+(\mathbb {R}) \cong H = \{ (x,y) \in \mathbb {R}^2 : y {\gt} 0\} \). We saw in Example 4.1.7 that the metric is given by \(g_{11} = \frac{1}{y^2} = g_{22}\), \(g_{12} = 0\), and we worked out the Levi-Civita connection in Example 4.3.10. Specifically, (4.13) implies that

$$\begin{align*} \Gamma _{11}^1 & = 0 & \Gamma _{12}^1 = \Gamma _{21}^1 & = -\frac{1}{y} & \Gamma _{22}^1 & = 0 \\ \Gamma _{11}^2 & = \frac{1}{y} & \Gamma _{12}^2 = \Gamma _{21}^2 & = 0 & \Gamma _{22}^2 & = -\frac{1}{y}. \end{align*}$$

Therefore, (4.15) becomes

$$\begin{align} \tag{4.16}\label{eq:hyperbolic geodesic coords} x''(t) - \frac{2}{y} x'(t)y'(t) & = 0 \\ y''(t) + \frac{1}{y}((x'(t))^2 - (y'(t))^2) & = 0. \nonumber \end{align}$$

On vertical rays, \(x(t) = \text{constant}\), so the first equation is satisfied and the second reduces to \(y''= \frac{(y')^2}{y}\), which has solution \(y(t) = c_1 e^{c_2t}\). So the vertical rays parametrized as \(\gamma (t) = (x_0,c_1e^{c_2t})\) are geodesics. We can eliminate the constants by deciding that we want \(\gamma (0) = (a,1)\), which implies \(c_1 = 1\), and that we want \(\gamma '(0) = (0,1)\):

\[ (0,1) = \gamma '(0) = (0,c_2) \]

so \(c_2 = 1\). That is, \(\gamma (t) = (a,e^t)\) is a geodesic in \(H\) for any \(a \in \mathbb {R}\).

As for the other geodesics, note that \(\frac{dy}{dx} = \frac{y'}{x'}\), which we can differentiate as follows:

\[ \frac{d^2 y}{dx^2} = \frac{d}{dx} \left( \frac{y'}{x'} \right) =\frac{x'y'' - y'x''}{(x')^2} \frac{1}{x'} \]

since \(\frac{df}{dx} = \frac{df}{dt} \frac{dt}{dx} = \frac{\frac{df}{dt}}{\frac{dx}{dt}}\). If we assume our curve \(\gamma (t) = (x(t),y(t))\) is a geodesic, we can substitute in \(x'' = \frac{2}{y} x'y'\) and \(y'' = \frac{1}{y} ((y')^2 - (x')^2)\) from (4.16):

$$\begin{align*} \frac{d^2 y}{dx^2} & = \frac{1}{(x')^3}\left(\frac{x'}{y}((y')^2-(x')^2) - \frac{2}{y}x'(y')^2\right) \\ & = \frac{1}{(x')^3}\frac{1}{y}\left(-(x')^3 - x'(y')^2\right) \\ & = -\frac{1}{y} \left(1 + \frac{(y')^2}{(x')^2}\right) \\ & = -\frac{1}{y} \left(1 + \left( \frac{dy}{dx}\right)^2\right). \end{align*}$$

(Note that we’re implicitly assuming \(x' \neq 0\) here, which is why we dealt with the \(x(t) = \text{constant}\) case separately.)

In other words,

\[ -1 = y \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = \frac{d}{dx} \left( y \frac{dy}{dx} \right) \]

on a geodesic. Integrating twice gives the solution

\[ x^2 + y^2 = ax + b, \]

which is the semi-circle of radius \(\sqrt{b+\frac{a^2}{4}}\) centered at the point \((a/2,0)\).

In other words, we’ve proved that the geodesics in \(H\) are the vertical rays and the semi-circles centered at points on the \(x\)-axis, as claimed in Example 2.3.22.

If we want to convert (4.15) to a system of \(2n\) first-order ODEs, we need to pass to the tangent bundle \(TM\). Recall that

Suppose \(p \in M\) lies in a coordinate chart \((U,\phi )\). If \(v \in T_p M\), then \(v = \sum _i y_i \frac{\partial }{\partial x_i}\) for some \(y_1, \dots , y_n\); then \((x_1, \dots , x_n, y_1, \dots , y_n)\) give local coordinates on \(T(\phi (U)) \cong U \times \mathbb {R}^n\); continuing down this path gives the proof of Exercise 1.2.6.

In any case, the curve \(t \mapsto \gamma (t)\) in \(M\) determines a curve \(t \mapsto (\gamma (t),\gamma '(t))\) in \(TM\) and, if \(\gamma \) is a geodesic, then in a local coordinate neighborhood \(T(\phi (U))\) this curve satisfies the first-order system

for all \(k =1 , \dots , n\).

By the usual existence and uniqueness theorems for solutions to ODEs, it will follow that, for any \(p \in M\) and any \(v \in T_pM\) (or, alternatively, any \((p,v) \in TM\)), there will be a geodesic \(\gamma \colon \thinspace \! (-\epsilon ,\epsilon ) \to M\) with \(\gamma (0) = p\) and \(\gamma '(0) = v\) and that any other solution will agree with \(\gamma \) on the overlap of their domains. We can package this up into a single vector field on \(TM\) which we will call the geodesic field.

Recall from Definition 1.7.1 that an integral curve of a vector field is one whose velocity agrees with the vector field at every point along the curve.

On the other hand, to prove that such a \(G\) exists, we can simply define it in coordinates by the system (4.17).

Example 4.4.7

Let \(M = S^1\) with its usual metric, namely \(g_\theta \left( \frac{\partial }{\partial \theta },\frac{\partial }{\partial \theta }\right) = 1\). It’s straightforward to check using (4.10) that

\[ \nabla _{\frac{\partial }{\partial \theta }}\frac{\partial }{\partial \theta } = 0. \]

So then the system (4.17) on \(TS^1 = S^1 \times \mathbb {R}\) with coordinates \((\theta , y)\) reduces to

$$\begin{align*} \theta '(t) & = y(t) \\ y'(t) & = 0. \end{align*}$$

Solutions are of the form \((\gamma (t),\gamma '(t)) = (c_1 + c_2 t,c_2)\), so when we require \(\gamma (0) = \theta _0\) and \(\gamma '(0) = y_0 \frac{\partial }{\partial \theta } \in T_{\theta _0}S^1\), we get \(\gamma (t) = \theta _0 + y_0 t\). These curves are trajectories of the field

\[ G(\theta ,y) = (y,0) \]

on the cylinder, as shown in Figure 4.8. This makes sense: geodesics go around the circle at constant speed, and this speed is determined by the initial velocity, which corresponds to height on the cylinder in this picture.

The existence of the geodesic flow follows from Proposition 1.7.2. More precisely, if \(U \subset M\) is an open set and \(\epsilon {\gt} 0\), we’ll use the notation

Then applying Proposition 1.7.2 to the geodesic field \(G\) on \(TM\) at the point \((p,0) \in TM\) yields:

Composing the local flow \(\phi \) with the projection \(\pi \colon \thinspace \! TM \to M\) yields the map \(\gamma = \pi \circ \phi \) which satisfies:

Notice that we can always make \(\delta \) bigger by making \(\epsilon \) smaller (or vice versa), so we can get \(\delta \) of uniform size if we want:

In particular, this means that, so long as \(v\) is short enough, \(\gamma (1,q,v)\) is well-defined. We should interpret \(\gamma (1,q,v)\) as follows: we travel along the geodesic through \(q\) in the direction of the unit vector \(\frac{v}{\| v\| }\) for a distance of \(\| v\| \). This idea of going out in the manifold in the direction indicated by \(v\) for a distance \(\| v\| \) is so useful that we give it a special name:

Most of the time, we will just be interested in the restriction of \(\exp \) to the tangent space at a point. That is, if \(B_\epsilon (0) \subset T_q M\) is the open ball of radius \(\epsilon \) centered at the origin in \(T_qM\), then we will define

by \(\exp _q(v) = \exp (q,v)\).

In the case that \(M\) is geodesically complete, meaning that every geodesic can be extended forever in both directions, then we can extent this map to the entire tangent space:

by definition of \(\gamma (t,q,v)\).

In other words, \(d(\exp _q)_0\) is the identity map on \(T_q M\), so the inverse function theorem (Theorem 1.4.7) implies that \(\exp _q\) is a local diffeomorphism.

4.5 Geodesics on Lie Groups

In Section 4.4 we defined geodesics as the curves \(\gamma \colon \thinspace \! I \to M\) with \(\frac{D\gamma '}{dt} = 0\), which means that geodesics are parametrized at constant speed and are straight (in the sense that they don’t turn). Compare to the fact that straight lines \(\alpha \colon \thinspace \! I \to \mathbb {R}^n\) traversed at constant speed satisfy \(\alpha ''(t)= 0\).

It seems intuitively obvious that the shortest path between two points should be as straight as possible, and hence should be a geodesic. Of course, we have to be careful because there is generally not a unique geodesic between two points (think about points on the sphere) and there may be no geodesic at all (think about the points \((2,0)\) and \((-2,0)\) in the annulus \(\{ (x,y) \in \mathbb {R}^2 : 1 {\lt} x^2 + y^2 {\lt} 4\} \)). But if two points are close enough, there will be a unique minimizing geodesic between them.

But, from Theorem 4.4.14, we know that for any \(p \in M\) there exists \(\epsilon {\gt} 0\) so that \(\exp _p \colon \thinspace \! B_\epsilon (0) \to M\) is a diffeomorphism. The image \(\exp _p(B_\epsilon (0))\) is called a normal ball or geodesic ball of radius \(\epsilon \) around \(p\), and denoted \(B_\epsilon (p)\).

The proof of this theorem is fairly technical and not very enlightening, so we skip it. For details, see do Carmo  [ 9 , § 3.3 ] .

Instead, let’s go a bit deeper on geodesics on Lie groups.

4.5.2 Geodesics in the Bi-Invariant Metric

Recall that we defined an exponential map \(\exp \colon \thinspace \! \mathfrak {g} \cong T_e G \to G\) for Lie groups in Section 3.10. We also defined a Riemannian exponential map \(\exp _e \colon \thinspace \! T_e G \to G\) in Definition 4.4.13. Of course, the Riemannian exponential depends on a choice of Riemannian metric, whereas the Lie group exponential map only depends on the Lie group structure. However, when we have a bi-invariant metric, the two exponential maps will agree, which is the main point of this section.

Example 4.5.10

Recall our left-invariant metric on \(\operatorname {Aff}^+(\mathbb {R}) \cong H\) from Example 4.1.7. We saw in Example 4.4.4 that the geodesics for this metric are vertical lines and semi-circles centered at points on the \(x\)-axis. Therefore, the geodesics through the identity \((0,1)\) are as depicted in Figure 4.9.

On the other hand, suppose \(v = a \frac{\partial }{\partial x} + b \frac{\partial }{\partial y} \in T_{(0,1)}H\) is a tangent vector at the identity. Under the identification of \(\operatorname {Aff}^+(\mathbb {R})\) with the matrix group \(\left\{ \begin{bmatrix} y & x \\ 0 & 1 \end{bmatrix} : (x,y) \in \mathbb {R}^2, y {\gt} 0\right\} \) from Example 3.2.3, \(v\) is identified with the \(2 \times 2\) matrix

\[ A_v = \begin{bmatrix} b & a \\ 0 & 0 \end{bmatrix} \in T_I \operatorname {GL}_2(\mathbb {R}). \]

In this setting, the exponential map is just the matrix exponential, so

\[ \exp (tv) = I + tA_v + \frac{1}{2!} (tA_v)^2 + \dots = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + t\begin{bmatrix} b & a \\ 0 & 0 \end{bmatrix} + \frac{t^2}{2!} \begin{bmatrix} b^2 & ab \\ 0 & 0 \end{bmatrix} + \frac{t^3}{3!} \begin{bmatrix} b^3 & ab^2 \\ 0 & 0 \end{bmatrix} + \dots = \begin{bmatrix} e^{tb} & \frac{a(e^{tb}-1)}{b} \\ 0 & 1 \end{bmatrix}. \]

In other words, the one-parameter subgroup associated to \(v \in T_{(0,1)}H\) is the curve \(t \mapsto \left(\frac{a(e^{tb}-1)}{b},e^{tb}\right)\), as depicted in Figure 4.11.

Clearly, the one-parameter subgroups are not geodesics with respect to the metric. The issue is, as we’ve already seen in Example 4.5.6, that \(\operatorname {Aff}^+(\mathbb {R})\) does not have a bi-invariant metric.

Theorem 4.5.11

In a bi-invariant metric on a Lie group, the one-parameter subgroups coincide with the geodesics through the identity.

Proof.

Recall from (4.10) the formula

\begin{equation*} g(\nabla _VU,W) = \frac{1}{2}\left[U(g(V,W)) + V(g(W,U)) - W(g(U,V)) - g([U,W],V) - g([V,W],U) - g([U,V],W)\right], \end{equation*}

Assume \(V,W \in \mathfrak {aff}^+(\mathbb {R})\) are left-invariant, and let \(U=V\) in the above equation. Since the metric is bi-invariant, \(g(V,W)\) and \(g(V,V)\) are constant, so the first three terms all vanish. The last term also vanishes since \([V,V]=0\), so we’re left with

\[ g(\nabla _VV,W) = \frac{1}{2}\left[- g([V,W],V) - g([V,W],V)\right] = -g([V,W],V) = g([W,V],V) = 0 \]

by Corollary 4.5.4.

Since \(W\) could have been any left-invariant vector field, this implies that \(\nabla _VV = 0\) for all left-invariant vector fields \(V\), where \(\nabla \) is the Levi-Civita connection. But then if \(\gamma (t) = \exp (tV_e)\) is the one-parameter subgoup associated with \(V_e \in T_eG\), we have that

\[ \gamma '(t) = \frac{d}{dt} \exp (tV_e) = (d L_{\exp (tV_e)})_e V_e = V_{\exp (tV_e)}, \]

so \(V\) is precisely the velocity field of the curve \(\gamma \), and hence

\[ 0 = \nabla _VV = \nabla _{\gamma '(t)} \gamma '(t) = \frac{D\gamma '}{dt} \]

implies that \(\gamma \) is a geodesic.

Since the Lie group exponential is defined in terms of one-parameter subgroups and the Riemannian exponential is defined in terms of geodesics, Theorem 4.5.11 implies that the Lie group exponential and the Riemannian exponential are the same on a Lie group with a bi-invariant metric.

In particular, this means that on matrix groups with a bi-invariant metric we can often solve the following problem: given \(h_1, h_2 \in G\), what is the minimizing geodesic from \(h_1\) to \(h_2\). To do so, note that it suffices to find a geodesic \(\gamma (t)\) from the identity element \(e\) to \(h_1^{-1}h_2\): since left-multiplication is an isometry, \(\alpha (t) = h_1 \gamma (t)\) will be a geodesic from \(h_1\) to \(h_2\).

So we’ve reduced to finding the geodesic from the identity to any group element \(h\). By Theorem 4.5.11, this geodesic will be a one-parameter subgroup passing through \(h\), so our goal is to find, say, a unit vector \(v \in T_e G\) so that \(\exp (t_0v) = h\). Then the curve \(\gamma \colon \thinspace \! [0,t_0] \to G\) given by \(\gamma (t) = \exp (tv)\) will be such a geodesic, and it will have length \(t_0\).

But inverting the matrix exponential is exactly what the matrix logarithm does. In general, since the log is multivalued, \(\log (h)\) will be a collection of elements of \(T_eG\), but if we choose \(u\) with smallest norm and let \(v = \frac{u}{\| u\| }\) then it will follow that \(d(e,h) = \| u\| \) and \(\gamma (t) = \exp (tv)\) is a minimizing geodesic from \(e\) to \(h\).

Example 4.5.12

Consider \(G = \operatorname {SO}(3)\). If \(A \in \operatorname {SO}(3)\), then \(A\) is unitarily diagonalizable:

\[ A = U D U^\ast , \]

where \(U \in \operatorname {U}(3)\) and \(D = \operatorname {diag}(1,e^{i\theta },e^{-i\theta })\) is the diagonal matrix with eigenvalues of \(A\) on the diagonal. Then

\[ \log (A) = U \log (D) U^\ast = U \operatorname {diag}(0,i\theta ,-i\theta ) U^\ast . \]

So then the distance from \(I\) to \(A\) is

\[ \| \log (A)\| = \sqrt{2}\theta \]

and the unit-speed geodesic from \(I\) to \(A\) is

\[ \gamma (t) = U \operatorname {diag}(1,e^{it/\sqrt{2}},e^{-it/\sqrt{2}}) U^\ast , \]

with \(A = \gamma (\sqrt{2}\theta )\).

Of course, the same sort of thing works on \(\operatorname {SO}(n)\) for any \(n\).

4.6 Geodesics on Grassmannians

The goal in this section is to describe geodesics on Grassmannians. This is knowledge that I have used many times in my research career; since Grassmannians pop up everywhere from string theory to random polygons to hyperspectral imaging to video analysis to algebraic geometry, you may find this information important at some point, too. The information in this section is mostly drawn from Edelman, Arias, and Smith’s amazing paper  [ 11 ] .

Recall from Section 3.12 that the Grassmannian \(\operatorname {Gr}(k,\mathbb {R}^n)\) is the collection of \(k\)-dimensional linear subspaces of \(\mathbb {R}^n\), and that it can be realized as a quotient of \(\operatorname {O}(n)\) by

In other words, we can interpret points in the Grassmannian as equivalence classes \([Q]\) of all orthogonal matrices whose first \(k\) columns span the same subspace as those of \(Q \in \operatorname {O}(n)\). Concretely,

The tangent space \(T_Q \operatorname {O}(n)\) splits into horizontal and vertical components: the vertical space is the subspace tangent to \([Q]\) (i.e., moving in those directions does not change the point on the Grassmannian), and the horizontal space is the orthogonal complement of the vertical space.

From the above description of \([Q]\), it should be pretty clear that the vertical space is

Hence, the horizontal space, which we can identify with \(T_{[Q]} \operatorname {Gr}(k,\mathbb {R}^n)\), consists of matrices of the form

where \(B\) is an arbitrary \((n-k) \times k\) matrix. Note that this immediately recovers the fact that \(\dim (\operatorname {Gr}(k,\mathbb {R}^n)) = k(n-k)\).

We get the standard Riemannian metric on \(\operatorname {Gr}(k,\mathbb {R}^n)\) by restricting the metric on \(\operatorname {O}(n)\) to the horizontal space (it is also traditional to multiply by \(1/2\) because of the repeated \(B\)’s in the expression for \(\Delta \) above). So what is the metric on the orthogonal group?

We get a left-invariant metric on \(\operatorname {O}(n)\) by pushing around any inner product we like on \(T_I\operatorname {O}(n) \subset T_I \operatorname {GL}_n(\mathbb {R}) = \operatorname {Mat}_{n \times n}(\mathbb {R})\), where the standard inner product is the Frobenius inner product

When we restrict this to \(T_I\operatorname {O}(n)\), which consists of the skew-symmetric \(n \times n\) matrices, we see that

using cyclic invariance of the trace and the fact that \(X^T = -X\). Therefore, Theorem 4.5.5 implies that the left-invariant metric \(g_{\operatorname {O}(n)}\) on \(\operatorname {O}(n)\) induced by \(\langle \cdot , \cdot \rangle _{\text{Fr}}\) is bi-invariant.

What is this metric? First, recall that, for \(Q \in \operatorname {O}(n)\), \(T_Q \operatorname {O}(n) = \{ Q X : X \text{ is skew-symmetric}\} \) since \(T_I \operatorname {O}(n)\) consists of skew-symmetric matrices and \((dL_Q)_I X = QX\). Then, for \(QX,QY \in T_Q \operatorname {O}(n)\),

Note that this is really just the Frobenius inner product of \(QX\) with \(QY\):

Therefore, with the factor of \(1/2\) mentioned previously, the induced metric on \(\operatorname {Gr}(k, \mathbb {R}^n)\) is given by, for \(\Delta _i = Q\begin{bmatrix} 0 & -B_i^T \\ B_i & 0 \end{bmatrix} \in T_{[Q]}\operatorname {Gr}(k,\mathbb {R}^n)\),

From Theorem 4.5.11 we know that geodesics on \(\operatorname {O}(n)\) take the form \(\gamma (t) = Q e^{tX}\) where \(Q \in \operatorname {O}(n)\), \(X\) is skew-symmetric, and we have \(\gamma (0) = Q\) and \(\gamma '(0) = QX \in T_Q \operatorname {O}(n)\).

If \(Q \in \operatorname {O}(n)\) and \(\Delta = Q\begin{bmatrix} 0 & -B^T \\ B & 0 \end{bmatrix} \in T_{[Q]}\operatorname {Gr}(k,\mathbb {R}^n)\), then

is a horizontal curve: the velocity vector \(\gamma '(t) = Q e^{t{\tiny \begin{bmatrix} 0 & -B^T \\ B & 0 \end{bmatrix}}} \begin{bmatrix} 0 & -B^T \\ B & 0 \end{bmatrix}\) is in the horizontal space at \(\gamma (t) = Qe^{t {\tiny \begin{bmatrix} 0 & -B^T \\ B & 0 \end{bmatrix}}}\). This means that \([\gamma (t)]\) is a geodesic in the Grassmannian, and conversely it can be shown that every geodesic in \(\operatorname {Gr}(k,\mathbb {R}^n)\) has a horizontal lift of this form.

In general, when we’re working with the Grassmannian, we’d rather avoid working with \(n \times n\) matrices (especially when \(k\) is much smaller than \(n\)). Since we also know that \(\operatorname {Gr}(k,\mathbb {R}^n) = \operatorname {St}(k,\mathbb {R}^n)/\operatorname {O}(k)\), we can also represent points in the Grassmannian by equivalence classes of \(n \times k\) matrices \(Y\) with orthonormal columns:

In practice, we can just think of \(Y\) as the first \(k\) columns of the orthogonal matrix \(Q\) from before. The above expression makes it clear that the vertical space of the projection \(\operatorname {St}(k,\mathbb {R}^n) \to \operatorname {Gr}(k,\mathbb {R}^n) = \operatorname {St}(k,\mathbb {R}^n)/\operatorname {O}(k)\) consists of \(n \times k\) matrices of the form

where \(A\) is \(k \times k\) skew-symmetric; that is, it corresponds to infinitesimal rotations of the basis for the subspace given by the columns of \(Y\). Therefore, the horizontal space, which we can identify with \(T_{[Y]} \operatorname {Gr}(k,\mathbb {R}^n)\), consists of all \(n \times k\) matrices \(H\) whose columns are perpendicular to the column space of \(Y\); that is,

Here’s another way to see this: if \(Y\) consists of the first \(k\) columns of an orthogonal matrix \(Q = \begin{bmatrix} Y & Z \end{bmatrix}\), then the first \(k\) columns of

are \(H = ZB\), and, since the columns of \(Y\) are perpendicular to the columns of \(Z\),

In terms of \(n \times k\) matrices, the Riemannian metric on the Grassmannian is still just the Frobenius inner product: if \(\Delta _i = Q\begin{bmatrix} 0 & -B_i^T \\ B_i & 0 \end{bmatrix}\) and \(H_i = ZB_i\) is the \(n \times k\) matrix consisting of the first \(k\) columns of \(\Delta _i\), then

If we’re working with \(n \times k\) matrices, we would then write

We can write geodesics in \(n \times k\) matrix format as well by just chopping off the last \(n-k\) columns; now our geodesics look like

where \(I_{n,k}\) consists of the first \(k\) columns of the \(n \times n \) identity matrix. Of course, it would be preferable to write this in terms of the \(n \times k\) matrices \(Y\) and \(H\) instead of the \(n \times n\) matrices \(Q\) and \(\begin{bmatrix} 0 & -B^T \\ B & 0 \end{bmatrix}\).

be the singular value decomposition of \(B\). Then

Therefore,

and hence (with some intermediate computations omitted),

But now \(H = ZB = (ZU_1)\Sigma V^T\) is the singular value decomposition of \(H\), so the result follows.

If \(H \in T_{[Y]}\operatorname {Gr}(k,\mathbb {R}^n)\) then, at least for small \(t\), Theorem 4.5.1 implies that the distance from \(\gamma (0) = Y\) to \(\gamma (t)\) is

where the \(\sigma _i\) are the singular values of \(H\); that is, the diagonal elements of \(\Sigma \).

If we write \(\theta _i = t \sigma _i\) and let \(\Theta = (\theta _1, \dots , \theta _k)\), then (4.19) becomes

which is the CS decomposition  [ 15 , 27 ] (see also Nick Higham’s blog post) of \(e^{t {\tiny \begin{bmatrix} 0 & -B^T \\ B & 0 \end{bmatrix}}} \in \operatorname {O}(n)\).

In the CS decomposition of a matrix, the \(\theta _i\) are the principal angles between the subspace spanned by the first \(k\) columns of the matrix and the coordinate subspace spanned by the first \(k\) standard basis vectors (i.e., the first \(k\) columns of the identity matrix). Moreover, the \(\cos \theta _i\) are the singular values of the \(k \times k\) Grammian \(I_{n,k}^T Y\), where \(Y\) is the \(n \times k\) matrix consisting of the first \(k\) columns.

More generally, if \(Y_1\) and \(Y_2\) are \(n \times k\) matrices with orthonormal columns, the cosines of the principal angles \(\theta _1, \dots , \theta _k\) between the subspaces they span are the singular values of the Grammian \(Y_1^T Y_2\).

In other words,

In fact, the singular value decomposition of \(Y_1^TY_2\) actually tells you what this geodesic is. Say

is the singular value decomposition. I can think of \(U\) and \(V\) as change of basis matrices on \(\operatorname {col}(Y_1)\) and \(\operatorname {col}(Y_2)\), respectively; that is, \([Y_1] = [Y_1U]\) and \([Y_2] = [Y_2V]\) and the columns of \(Y_1U\) and \(Y_2V\) give new orthonormal bases for the these respective subspaces (i.e., points on the Grassmannian). If I let \(u_i\) be the \(i\)th column of \(Y_1U\) and let \(v_j\) be the \(j\)th column of \(Y_2V\), then

This tells us that the bases \(u_1, \dots , u_k\) for \([Y_1]\) and \(v_1, \dots , v_k\) for \([Y_2]\) are in some sense optimally aligned: in particular, they realize the principal angles between the subspaces, since the angle between \(u_i\) and \(v_i\) is \(\theta _i\). This is sketched in Figure 4.12.

And now, the geodesic from \([Y_1]\) to \([Y_2]\) is simple:

For example, \(\operatorname {Gr}(1,\mathbb {R}^n) = \mathbb {RP}^{n-1}\) consists of lines through the origin, and this tells us that the geodesic between two lines is just the steady-speed rotation of one to the other.

Example 4.6.4

In general, I can get random points in \(\operatorname {Gr}(k,\mathbb {R}^n)\) by applying Gram–Schmidt to \(n\times k\) matrices with standard Gaussian entries.

With \(k=26\) and \(n=57\) I generated \(Y_1\) and \(Y_2\) in this way; they’re visualized in the first row of Figure 4.14 by having the value in each entry specify a color. The second row of Figure 4.14 shows the aligned representatives \(Y_1U\) and \(Y_2V\). Using Theorem 4.6.2, \(d([Y_1],[Y_2]) \approx 4.692\).

    

    

In turn, Figure 4.15 shows a few equally-spaced points along the minimizing geodesic given by Theorem 4.6.3 connecting these two points in the Grassmannian.

                 

Here is a video showing the full geodesic:

Using Theorem 4.6.2 it is straightforward to compute distances between points in the Grassmannian. Figure 4.16 shows the histogram of distances between 10,000 pairs of random points in \(\operatorname {Gr}(26,\mathbb {R}^{57})\), which fits quite well to the Gaussian distribution with mean \(4.844\) and standard deviation \(0.0788\). Note that the maximum possible distance is \(\sqrt{26}\frac{\pi }{2}\approx 8.0095\).

4.7 The Curvature Tensor

Given the name, you might expect that some of the key ideas in Riemannian geometry are due to Bernhard Riemann. Indeed, he explained a way of thinking about what eventually came to be called Riemannian manifolds and curvature in his Habilitationsschrift from 1854  [ 28 ] .

His idea for how to define curvature in a manifold was something like the following. Let \(p \in M\) and consider \(\sigma \subset T_pM\), a 2-dimensional subspace. Each line through the origin in \(\sigma \) exponentiates out to give a geodesic in \(M\), and by taking all such geodesics we get a surface \(S\) in \(M\) containing \(p\) and tangent to \(\sigma \) at \(p\). The Riemannian metric on \(M\) induces a Riemannian metric on \(S\).

Gauss had showed in 1827  [ 13 ] that the Gauss curvature of a surface depends only on its First Fundamental Form (a.k.a., Riemannian metric); Riemann called this the curvature at \(p\), denoted \(K(p,\sigma )\). These days we call this curvature the sectional curvature at \(p\) with respect to \(\sigma \).

That said, Riemann did not give any way to compute this curvature and it took quite a long time to get a rigorous, modern definition. Unfortunately, Riemann’s intuition, which hopefully makes sense, is not really evident in the modern definition, which seems very abstract.

That said, I’m going to give you the modern definitions and try to connect it back to the geometric intuition as much as I can.

Despite having called it the curvature tensor, I have not yet showed that \(R\) is a tensor field. To be a proper tensor field, the value of \(R(X,Y)Z\) at a point would need to depend only on the values of \(X,Y,Z\) at the point; given all the derivatives involved, you might expect that it would depend on the values of \(X,Y,Z\) in a neighborhood of the point.

Of course, (4.21) shows that \(R(X,Y)Z\) is multilinear over \(\mathbb {R}\).

Hence, \(R(X,Y)Z\) is multilinear in \(X\) over \(C^\infty (M)\). Since \(R(X,Y)Z = -R(Y,X)Z\), this shows that it is also multilinear in \(Y\) over \(C^\infty (M)\).

Finally,

showing the linearity of \(R(X,Y)Z\) in \(Z\) over \(C^\infty (M)\) and completing the proof.

and

Therefore,

and similarly

If we add (4.26), (4.27), and (4.28), all the connection terms cancel out and we’re left with

by the Jacobi identity Proposition 1.6.43.

We’ve defined \(R\) as a map \(\mathfrak {X}(M) \times \mathfrak {X}(M) \times \mathfrak {X}(M) \to \mathfrak {X}(M)\); that is, \(R(X,Y)Z \in \mathfrak {X}(M)\). We can interpret this as a \((1,3)\)-tensor field as follows: at each \(p \in M\) we get a map \((T_pM)^\ast \times T_pM \times T_pM \times T_pM \to \mathbb {R}\) given by

(Note that this being well-defined depends essentially on Exercise 4.7.10.)

Of course, we’ve defined everything in the presence of a Riemannian metric \(g\), and we can always convert vectors into linear functionals with inner products: \(T \leftrightarrow g(\cdot , T)\) for any \(T \in \mathfrak {X}(M)\), and so we can convert \(R\) into a \((0,4)\)-tensor field—that is, a \(C^\infty (M)\)-multilinear map \(\mathfrak {X}(M) \times \mathfrak {X}(M) \times \mathfrak {X}(M) \times \mathfrak {X}(M) \to \mathbb {R}\) by mapping

2 follows from the fact that \(R(X,Y) = -R(Y,X)\), which we saw was a direct consequence of the definition.

For 3, observe that

since \(XY-YX = [X,Y]\). Now, if we repeatedly use Corollary 4.3.5 we see that

Subtracting the second and third lines from the first yields

which implies 3.

For 4, note that the Bianchi identity implies that

Adding the first two lines and subtracting the third and fourth and using 2 and 3 to cancel four pairs of terms and simplify the remaining terms yields

which implies the result.

In local coordinates, if we write \(X_i = \frac{\partial }{\partial x_i}\), then there exist smooth functions \(R_{ijk}^s\) so that

The functions \(R_{ijk}^s\) are usually called the components of \(R\) in the given local coordinates. If we have vector fields \(U, V, W\) with local coordinate expressions

(note that now I am being more careful about whether indices are superscripted or subscripted), then

(In Einstein notation, this gets written as \(R(U,V)W = R_{ijk}^s u^j v^k w^i X_s\), where summation always happens over repeated indices, and repeated indices should appear in up/down pairs.)

When we lower an index and think of \(R\) as a \((0,4)\)-tensor, then we get components

Then Proposition 4.7.12 implies the following symmetries for the \(R_{ijks}\):

In particular, recall that we proved the interchange symmetry 4 from the other three, and the first three turn out to be a complete list of symmetries of the curvature tensor, so \(R\) is completely specified in local coordinates by the choice of \(\frac{n^2(n-1)^2}{12}\) independent functions \(R_{ijks}\).

4.8 Sectional Curvature

We are now in a position to define sectional curvature, which will agree with Riemann’s original notion of curvature. To do so, we introduce one piece of notation: for \(V\) an inner product space and \(u,v \in V\), let \(\| u \wedge v\| \) be the area of the parallelogram spanned by \(u\) and \(v\); that is,

Since the right hand side involves a choice of basis for \(\sigma \), it is not at all clear yet that \(K(\sigma )\) is well-defined.

Then \(K(u,v)\) has the following symmetries:

1. \(K(u,v) = K(v,u)\) by skew-symmetry of \(R\) (Proposition 4.7.122 and 3) and inspection of (4.29).

2. \(K(u,v) = K(\lambda u, v)\) for any \(\lambda \in \mathbb {R}\) by multilinearity of \(R\) and homogeneity of \(\| u \wedge v\| \).

3. \(K(u,v) = K(u + \lambda v, v)\) by the multilinearity and skew-symmetry of \(R\) and \(\wedge \).

The transformations \((u,v) \mapsto (v,u)\), \((u,v) \mapsto (\lambda u, v)\), and \((u,v) \mapsto (u + \lambda v, v)\) correspond to the \(2 \times 2\) matrices \(\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\), \(\begin{bmatrix} \lambda & 0 \\ 0 & 1 \end{bmatrix}\), and \(\begin{bmatrix} 1 & \lambda \\ 0 & 1 \end{bmatrix}\), respectively. These matrices generate \(\operatorname {GL}_2(\mathbb {R})\), so we’ve just shown that \(K\) is invariant under the \(GL_2(\mathbb {R})\) action on \(\sigma \). Of course, \(\operatorname {GL}_2(\mathbb {R})\) transforms any basis into any other basis, so this shows that \(K\) is basis-independent.

4.9 Ricci and Scalar Curvatures

Sectional curvature is an extremely powerful tool for describing the geometry of a manifold, and there is a huge amount of work out there about the impact it has on dynamics, the constraints it places on topology, and many other interactions it has with other concepts of interest. Just to give one example, the sphere theorem says that any complete, simply-connected \(n\)-manifold with sectional curvatures in the interval \([1,4)\) is homeomorphic to the sphere \(S^n\). This was conjectured by Hopf in 1932  [ 19 ] and proved independently in 1961 by Berger  [ 5 ] and Klingenberg  [ 21 ] . In 2007, Brendle and Schoen  [ 6 ] took this further and proved the differentiable sphere theorem, which says that any Riemannian manifold with these sectional curvature bounds is diffeomorphic to \(S^n\) with its standard smooth structure. These bounds are optimal: the complex projective space \(\mathbb {CP}^n\) with the Fubini–Study metric has sectional curvatures in the closed interval \([1,4]\), and is not homeomorphic to a sphere unless \(n=1\).

That said, while it’s conceptually somewhat simpler than the full Riemann curvature tensor, sectional curvature is still rather complicated. One way to simplify it is to take averages in the following sense: suppose \(p \in M\) and \(X \in T_pM\). Then we could take the average of all sectional curvatures of 2-planes \(\sigma \) containing \(X\). Of course, this really only depends on the line spanned by \(X\), so this assigns a number to tangent lines at points. Indeed, this is the basic idea of Ricci curvature.

In other words, it’s enough just to average over any orthonormal basis and this will give the same answer as averaging over every 2-plane containing \(X\). In fact, for reasons that will become apparent when we discuss this a little more abstractly, most people don’t actually take averages, they just sum over orthonormal bases:

By Exercise 4.9.2, \(\frac{1}{n-1} \operatorname {Ric}_p(X)\) is the average (in either sense) of the sectional curvatures of all 2-planes containing \(X\). Of course, there’s no reason to stop averaging there. What if we take the average of Ricci curvatures over all directions? Again, we’re going to sum rather than averaging, but this gives scalar curvature:

Now we make things a bit more abstract. By definition, if \(X \in T_pM\) is a unit vector, then we can complete to an orthonormal basis \(Z_1, \dots , Z_{n-1}, Z_n = X\) in any way we like and we have

Writing \(K(X,Z_i) = g(R(X,Z_i)Z_i,X)\) (since \(\| X \wedge Z_i\| ^2=1\) for all \(i\)), we can also write this as

In fact, since skew-symmetry of \(R\) implies that \(R(X,Z_n) = R(X,X) = 0\), we can sum from 1 to \(n\) instead:

In other words, \(\operatorname {Ric}_p(X)\) is the trace of the linear map \(T_p M \to T_p M\) given by

In particular, since trace is the sum of the diagonal entries in any matrix representation of a linear transformation, this shows that the formula (4.34) for \(\operatorname {Ric}_p(X)\) is valid for any orthonormal basis \(Z_1, \dots , Z_n\) for \(T_pM\): \(Z_n\) is not required to equal \(X\).

Now, if we forget about the requirement that \(X\) be a unit vector, the expression

is a quadratic form, which we can polarize to get a symmetric bilinear form

which we’ve written here in terms of an orthonormal basis. Again, this is a trace so we can write more abstractly.

Let’s recall how to compute the trace of a linear transformation of \(T_pM\). Say \(X_1, \dots , X_n\) is a basis for \(T_pM\), and as usual write \(g_{ij} := g_p( X_i, X_j)\). If \(A \colon \thinspace \! T_pM \to T_pM\) is linear, with matrix \(\begin{bmatrix} a_j^i \end{bmatrix}_{i,j}\) with respect to our chosen basis, then that means

so we have

Therefore, if the inverse matrix of \(\begin{bmatrix} g_{ij} \end{bmatrix}_{i,j}\) is \(\begin{bmatrix} g^{k\ell } \end{bmatrix}_{k,\ell }\), then we have

In particular, then, if \(X_i = \frac{\partial }{\partial x_i}\) in some local coordinate chart around \(p\), then this implies that

This is symmetric, so we can also write \(\operatorname {Ric}_{ij}(p) = \sum _k R_{ikj}^k\) or, in abstract index notation (with the Einstein summation convention),

that is, the Ricci tensor is the contraction of the Riemann curvature tensor in the first and third indices.

4.10 Gradients and Optimization

In practical problems, we are very often interested in minimizing (or maximizing) some function \(f \colon \thinspace \! M \to \mathbb {R}\).

Of course, the most naïve way of minimizing a function is to do gradient descent: that is, to flow in the direction of the negative gradient. But that raises the question: what is the gradient on a manifold, and does gradient descent still make sense?

I have already implicitly addressed the first question in Section 2.2, but let’s make that discussion clearer and more explicit. The first thing to do is to recall a useful characterization of the gradient of a function \(f \colon \thinspace \! \mathbb {R}^n \to \mathbb {R}\): for a tangent vector \(v\),

that is, the dot product of the gradient of \(f\) with any tangent vector \(v\) gives the directional derivative of \(f\) in the direction of \(v\).

We now have all the machinery in place to interpret everything in Equation 4.35 on manifolds. For the left hand side, we should think of \(\nabla f(p)\) and \(v\) as elements of \(T_pM\), and then a Riemannian metric \(g\) tells us the inner product on \(T_pM\), so the left-hand side becomes \(g_p(\nabla f(p), v)\). In fact, I’m going to use \(\operatorname {grad}f\) rather than \(\nabla f\) for the gradient on manifolds, since we’re already using \(\nabla \) for the Levi-Civita connection, so the left-hand side is really going to be \(g_p(\operatorname {grad}f(p), v)\).

For the right-hand side of Equation 4.35, recall that way back in Section 1.2 we defined tangent vectors as directional derivative operators, so the right way of saying \(\frac{\partial f}{\partial v}\) is \(v(f)\).

In other words, we can get a generalization of Equation 4.35 to arbitrary Riemannian manifolds by writing

In fact, if we recall Lemma 1.3.4, we know that \(v(f) = df_p(v)\) is just the differential of \(f\) evaluated on \(v \in T_p M\). So we can also write

for any \(v \in T_pM\). This will be, in fact, the definition of the gradient we give below in Definition 4.10.2, but let’s put this in a slightly more general framework first (building on the informal discussion in Section 2.2).

Say that \(f \colon \thinspace \! M \to \mathbb {R}\) is a smooth function. Then the differential \(df \in \Omega ^1(M)\) is a 1-form. Moreover, if \(M\) is not just a manifold but a Riemannian manifold, then I can use the Riemannian metric to identify 1-forms with vector fields.

More precisely, for any 1-form \(\omega \in \Omega ^1(M)\), define \(\omega ^\sharp \in \mathfrak {X}(M)\) by

for any \(v \in T_pM\). For each \(p \in M\) we know there is such a vector \(\omega ^\sharp (p) \in T_pM\) by the Riesz Representation Theorem, and the fact that \(\omega ^\sharp \) is smooth follows from the smoothness of \(g\) and \(\omega \).

More abstractly, we could write

In local coordinates, we have that \(\omega = \sum _i w_i dx^i\) and any \(v = \sum _i v^i \frac{\partial }{\partial x_i}\), so

On the other hand, \(\omega ^\sharp = \sum _i u^i \frac{\partial }{\partial x_i}\) and

Equating coefficients gives

or

Putting this together, then, we have that for \(\omega = \sum _i w_i dx^i\), the dual vector field is

When we let \(\omega = df\), the resulting vector field is the gradient:

If we just unwind the definitions and recall Lemma 1.3.4, we see that, for any \(X \in \mathfrak {X}(M)\),

as in (4.36) and (4.37).

In coordinates, we know that

so (4.38) tells us that

In fact, this is true more generally:

by Proposition 2.6.4. Therefore, for any \(v \in T_pN\),

Of course, \(d\iota _p \colon \thinspace \! T_pN \to T_p M\) is just the inclusion map so

The only vector which satisfies this equation for all \(v \in T_pN\) is the orthogonal projection of \(\overline{\operatorname {grad}f}\) onto \(T_pN\).

In particular, when \(N \subset \mathbb {R}^n\) is a smooth submanifold, then the Riemannian gradient \(\operatorname {grad}f\) at a point \(p\) is simply the orthogonal projection of the usual Euclidean gradient \(\nabla f\) onto \(T_pN\). In many cases (spheres, Stiefel manifolds, etc.) we are interested in smooth submanifolds of Euclidean space and so we can compute gradients in this fairly straightforward way.

Note that the Riemannian gradient has the same interpretation as the direction of steepest ascent:

with equality if and only if \(\operatorname {grad}f\) and \(v\) point in the same direction.

When \(\operatorname {grad}f\) and \(v\) point in the same direction,

since \(\| v\| = 1\).

Also, just as in Euclidean space, the gradient of a function is perpendicular to level sets of the function.

For \(f \colon \thinspace \! M \to \mathbb {R}\) smooth, we’ve now seen that \(\operatorname {grad}f \in \mathfrak {X}(M)\). Then the gradient flow of \(f\) is simply the local flow of the vector field \(\operatorname {grad}f\). We can also pose this as an ODE:

Bringing this back to Example 4.10.1, Dustin Mixon, Tom Needham, Soledad Villar and I  [ 25 ] proved that for almost every \(F_0 \in M\), the negative gradient flow of \(\operatorname {FP}\) starting at \(F_0\) exists for all time and limits, as \(t \to +\infty \), to a unit-norm tight frame; that is, a global minimizer of \(\operatorname {FP}\). This is true in spite of the fact that the frame potential \(\operatorname {FP}\) is not convex: indeed, it has lots of non-minimizing critical points.

Numerically, how can we actually do gradient descent? Well, at each point \(p\) we want to move in the direction of the vector \(-\operatorname {grad}f(p)\). But we know how to do this (at least in principle): use the Riemannian exponential map! So an algorithm for Riemannian gradient descent looks something like this:

Of course, computing the Riemannian exponential map generally involves solving the geodesic equation, which is a second-order ODE, so as stated Algorithm 1 is not usually very practical. The typical strategy for doing gradient descent in practice is to find some approximation to the exponential map which is well-adapted to the manifold of interest.

For example, if \(M \subset \mathbb {R}^n\) is a smooth submanifold, then for small \(\gamma {\gt} 0\), \(\exp _p(\gamma v)\) will be very close to the point in \(M\) which is closest to \(p + \gamma v \in \mathbb {R}^n\). Hence, if there is a good way of projecting a point in \(\mathbb {R}^n\) back to \(M\), we can replace \(\exp _{p_n}(-\gamma \operatorname {grad}f(p_n))\) in Algorithm 1 with the result of applying this projection to \(p_n - \gamma \operatorname {grad}f(p_n) \).

Example 4.10.11

Consider \(\operatorname {SO}(n) \subset \operatorname {Mat}_{n \times n}(\mathbb {R})\). Then for a function \(f \colon \thinspace \! \operatorname {SO}(n) \to \mathbb {R}\) with a smooth extension to (a neighborhood of \(\operatorname {SO}(n)\) in) \( \operatorname {Mat}_{n \times n}(\mathbb {R})\), we can approximate the gradient descent of \(f\) as in Algorithm 2 .

This depends on two functions, ProjSOn and PolarDecomp, defined as follows.

• For \(A \in \operatorname {SO}(n)\) and \(X \in \operatorname {Mat}_{n \times n}(\mathbb {R})\), ProjSOn\((A,X)\) is the orthogonal projection of \(X\) to

  \[ T_A \operatorname {SO}(n) = \{ A \Delta : \Delta \in \operatorname {Mat}_{n \times n}(\mathbb {R}) \text{ is skew-symmetric}\} . \]

• For \(X \in \operatorname {Mat}_{n \times n}(\mathbb {R})\), PolarDecomp\((X)\) returns the pair \((U,P)\), where \(X = UP\) is the polar decomposition of \(X\). Note that \(U\) is the closest matrix in \(\operatorname {SO}(n)\) to \(X\). ⁴⁰